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Algebra II Final Exam Review Notes (Transcript-based)

Question 1: Inverse variation

In inverse variation, the relationship between two variables is of the form y= rac{a}{x} where the constant of variation is denoted by a (some books use k). Because the variables vary inversely, the denominator contains the variable x. Given that when x=-8 we have y=3, substitute to find the constant:
[3pt] 3= rac{a}{-8} \Rightarrow a=-24. ]
So the equation relating the variables is y= rac{-24}{x}. Then Part B asks to find y when x=4:
[3pt] y= rac{-24}{4}=-6. ]
Key takeaway: for a two-variable inverse variation, use y= rac{a}{x} and determine a from any given pair (x,y), then substitute to find the desired value of y.

Question 2: Joint variation

If a variable varies jointly with two others, the relationship is typically of the form z=a\,x\,y (product of the two independent variables with a constant of variation). Given z=-72 when x=9 and y=-4, substitute:
[3pt] -72=a\,(9)(-4)=-36a \Rightarrow a=2. ]
Thus the equation is z=2xy.
Part B asks for z when x=6 and y=4:
[3pt] z=2(6)(4)=48. ]
Summary: for joint variation (with two variables multiplied together), find the constant from a known data point, then use the specific equation to compute new values.

Question 3: Combined variation

Combined variation combines direct variation with inverse variation in a single expression.

  • Part A: d varies directly with m and inversely with the cube of p. This yields d= rac{a m}{p^{3}} where a is the constant of variation. (If a data point is given, solve for a; otherwise present the general form.)
  • Part B: r varies jointly with the variable y and the square root of x. This yields r=a\,y\sqrt{x}. (Again, determine a from given data, or state the form.)

Question 4: Graphing a transformed reciprocal function

Given y=-\frac{2}{x-1}+3, compare to the parent function y= rac{a}{x} (here a=-2). The transformation consists of horizontal shift by h=1 to the right, and vertical shift by k=3 up. The key features:

  • Parent: y=-\frac{2}{x} shapes the hyperbola.
  • Horizontal asymptote: the vertical shift moves it up, giving y=3 as the horizontal asymptote.
  • Vertical asymptote: the horizontal shift moves the vertical asymptote to x=1.
    Thus asymptotes: x=1 and y=3.

Domain and range:

  • Domain: (-\infty,1)\cup(1,\infty) (we exclude the vertical asymptote at x=1).
  • Range: (-\infty,3)\cup(3,\infty) (we exclude the horizontal asymptote at y=3).

How to sketch (conceptually): shift the origin to the new centered point around the asymptotes, and use the parent function y=-\frac{2}{x} to place points. Example points relative to the shifted origin: left 2, up 1; left 1, up 2; and similarly on the right side. The branches approach the asymptotes but never touch.

Question 5: Rational expressions; simplification and domain exclusions

Part A: Simplify \frac{x^{2}-16}{x^{2}-7x+12}.

  • Recognize difference of squares in the numerator: x^{2}-16=(x+4)(x-4).
  • Factor the denominator: x^{2}-7x+12=(x-3)(x-4).
  • Cancel the common factor x-4 to obtain \frac{x+4}{x-3}.
  • Excluded values (before cancellation): values that make any original denominator zero, i.e. x\neq 3,4. After cancellation, the reduced expression remains defined for all other x except those two.
    Final: \frac{x+4}{x-3},\quad x\neq 3,4.

Part B: Simplify \frac{5x+20}{3x^{2}+11x-4}.

  • Factor common factor in the numerator: 5(x+4).
  • Factor the denominator: find numbers multiplying to 3(-4)=-12 that sum to 11. These are 12 and -1, giving the factorization 3x^{2}+11x-4=(3x-1)(x+4).
  • Cancel the common factor x+4 to get \frac{5}{3}(x-1).
  • Excluded values: from the original denominator, x\neq -4,\ x\neq \frac{1}{3}. The simplified expression is \frac{5}{3}(x-1) with those exclusions.

Part C: Division of polynomials (a fraction divided by another fraction). The idea is to multiply by the reciprocal after factoring. Starting from a given complex rational expression, factor completely, then invert and multiply. After full factorization and cancellation, the result simplifies to a polynomial or a rational expression with cancelled factors removed; check exclusions by inspecting the original denominators before cancellation.

Question 6: Subtracting fractions with polynomial denominators

We want a common denominator for (\frac{2}{x-5} - \frac{2}{x+4}). The common denominator is ((x-5)(x+4)). Multiply each fraction to obtain equal denominators:
[\frac{2}{x-5} = \frac{2(x+4)}{(x-5)(x+4)},\quad \frac{2}{x+4}=\frac{2(x-5)}{(x-5)(x+4)}.]
Subtract numerators:
[\frac{2(x+4) - 2(x-5)}{(x-5)(x+4)} = \frac{2x+8 - 2x + 10}{(x-5)(x+4)} = \frac{18}{(x-5)(x+4)}.]
So the result is \frac{18}{(x-5)(x+4)}, with the understanding that x\neq 5,-4 in the original expression (these would make a denominator zero).

Question 7: Solving equations (proportions and rational equations)

  • Part A (a proportion): Solve \frac{3}{2x-1}=\frac{1}{2x-1}. Cross-multiplication yields a linear equation; the example shows solving by cross-multiplication and simplifying to get x=5, while noting the extraneous-solution caveat: you must check that you didn’t create a division by zero (i.e., you must avoid denominators equal to zero).
  • Part B: Clear denominators by multiplying every term by a common denominator (here, 18). After distributing and simplifying, you may obtain a simple linear equation; in the example, solving gives x=-\frac{1}{3}. Always check for extraneous solutions by ensuring denominators are not zero.
  • Part C: Solve a more complex rational equation by obtaining a common denominator across all terms, clearing, and solving the resulting polynomial equation. You may obtain two potential solutions; check which remain valid by ensuring none makes any denominator zero (i.e., no extraneous solutions). In the example, you get x=0 or x=2, and you verify that neither makes a denominator zero (excluded values would be x=-3 or x=3 in the setup).

Question 8: Distance and midpoint between two points

Part A: Distance between ((3,-8)) and ((-4,16)): use
[d=\,\sqrt{(x2-x1)^2+(y2-y1)^2}].
Here, (x2-x1=-4-3=-7) and (y2-y1=16-(-8)=24). So
[d=\sqrt{(-7)^2+24^2}=\sqrt{49+576}=\sqrt{625}=25.]
Part B: Midpoint: use
[M=\left(\frac{x1+x2}{2},\frac{y1+y2}{2}\right)].
Thus, (M=\left(\frac{3+(-4)}{2},\frac{-8+16}{2}\right)=\left(-\frac{1}{2},4\right).]
So distance = 25; midpoint = ((-0.5,\ 4)).

Question 9: Basic probability

There are 7 blue, 3 red, 5 white marbles: total = 15. Probability of blue is
[P(B)=\frac{7}{15}].

Question 10: Permutations and factorials

With six distinct books, the number of possible orders on a shelf is the number of permutations of 6 objects:
[6! = 720].

Question 11: Three-number combinations from 0–9 without repetition

Consider three positions on a combination lock (no repetition). The transcript explains this as a combination problem (order does not matter): NCR with N=20 candidates and r=2 spots. The explicit calculation given is
[\binom{20}{2} = \frac{20!}{(20-2)!\,2!}=\frac{20\cdot 19}{2}=190.]
For the numeric reasoning about the first digit, second digit, etc., the speaker notes 10 choices for the first digit (0–9), then 9 for the second, then 8 left for the third; however, the transcript reports a particular counting case yielding 450 by constraining the last digit to be odd and the first digit not to be zero. The key combinatorial takeaway is the distinction between permutations (order matters) and combinations (order does not). The standard approach for “no repetition” three-digit selections from 0–9 is either permutations (10P3) or combinations (10C3) depending on whether order matters; the transcript specifically asserts the combinatorial (order not mattering) interpretation in this problem and provides the numeric result 450 for their chosen constraints.

Question 12: Combinations vs permutations; two identical jobs

Two spots to fill from 20 candidates; the jobs are identical, so order does not matter. This is a combinations problem:\binom{20}{2}=190. The transcript also notes the distinction between permutations ($NPR$) and combinations ($NCR$).

Question 13: Distinguishable permutations of words

  • For the word "algebra": there are 7 letters with two As identical. The number of distinguishable permutations is
    [\frac{7!}{2!}=\frac{5040}{2}=2520.]
  • For the word "Mississippi": 11 letters with counts S=4, I=4, P=2, and M=1. The number of distinct permutations is
    [\frac{11!}{4!\,4!\,2!}].] (The transcript notes the idea and the factorial simplifications such as 4! = 24, but the final computation is left in factorial form or evaluated as needed.)

Question 14: Finishing a track race; order matters

If 10 runners finish first, second, and third, the order matters (gold, silver, bronze). This is a permutation problem: 10P_3=\frac{10!}{(10-3)!}=10\cdot9\cdot8=720. The transcript also relates this to the intuitive product rule: 10 choices for first, then 9 for second, then 8 for third.

Question 15–16: Binomial expansion and coefficients

  • Coefficient of x^4 in the expansion of ((2x-3)^7): Use the general term for binomial expansion: for ((a+b)^n), the term with x^k arises from choosing k copies of the part containing x. Here, to get x^4 we need k=4, so the coefficient is
    [\binom{7}{4}(2x)^{4}(-3)^{3} = 35 \cdot 16 \cdot (-27) = -118…20.]
    This yields the term (-15{,}120\,x^{4}) (i.e., the coefficient is (-15{,}120)).

  • The transcript also discusses using Pascal’s triangle to locate specific terms without full expansion. For example, the third term in ((x+y)^3) corresponds to (\binom{3}{2}x^{1}y^{2}=3xy^{2}) and similarly for other rows; the general rule is that the term indices align so that the exponents add to the overall exponent and the binomial coefficients follow the row number.

  • An additional example from the transcript notes the term in ((x+2)^3), where the third term is (\binom{3}{2}x^{1}2^{2}=3\cdot x\cdot 4=12x). The key idea is that the nth row (starting from n=0) of Pascal’s triangle provides the binomial coefficients for the expansion.

  • Binomial coefficient general formula: For ((a+b)^n), the k-th term is
    [\binom{n}{k-1}a^{n-(k-1)}b^{k-1}.
    ] This provides a direct route to the coefficient of any specific power of a or b without full expansion.

  • Binomial probability (example): If you want the probability of exactly 7 successes in 10 trials with success probability p=0.8, use
    [P(K=7)=\binom{10}{7} p^{7} (1-p)^{3}.
    ] The transcript notes the calculation using the formula and reports approximately 0.20 (20%). If you wanted “seven or more,” you would sum the probabilities for 7, 8, 9, 10; for seven or fewer, you would sum from 0 to 6.

Question 18: Descriptive statistics; mean, median, mode, range, standard deviation, outlier

  • Given quiz scores: 10, 16, 6, 16, 5, 8 (actual values in the transcript show a later correction; we’ll present the intended logic and final results).
  • Mean: sum divided by count; the transcript arrives at a mean of 8 (with a later correction confirming the sum they used). The corrected mean from the set provided is 8.
  • Median: when data are ordered, the middle value or the average of the two middle values. If the data set has two middle values, the median is their average; here the two middle values yield a median of 7 in the transcript’s example.
  • Mode: the most frequent value. The transcript identifies two modes: 5 and 8 (two values tied as most frequent).
  • Range: max minus min; the transcript gives 16 − 5 = 11.
  • Outlier: the value farthest from the rest; the transcript identifies 16 as an outlier in this example.
  • Standard deviation (by hand): compute deviations from the mean, square, sum, divide by the number of data points, then take the square root. The transcript shows a hand-worked example yielding approximately \sigma\approx 3.8.
  • Part B (shift by +1): adding 1 to every data point increases the mean to 8+1=9; the standard deviation remains the same because spread is unchanged. If you multiply all values by a constant, all descriptive statistics scale accordingly (mean, median, mode, range, standard deviation all multiply by the same constant); if you add a constant to all values, the mean, median, and mode all shift by that constant, while the standard deviation and range do not change.

Question 19: Standard normal distribution and z-scores

  • Given mean height 5'6" (66 inches) and standard deviation 3 inches; probability a randomly selected student is over 6'0" (72 inches):
    [z=\frac{72-66}{3}=2.]
  • The probability of being above two standard deviations above the mean is about 0.025 (2.5%), according to standard normal tables or a calculator. The transcript notes this region as the area to the right of z=2.

Question 20: Margin of error for a survey

  • The transcript uses the simplified formula for margin of error:
    [\text{MOE} = \pm\sqrt{\frac{1}{n}}.]
  • For a survey of n=200 with 30% saying they bring their lunch, the MOE is approximately
    [\sqrt{\frac{1}{200}}\approx 0.0707\approx 7\%.
  • Interpretation: the true proportion lies within about ±7 percentage points of the observed 30% (i.e., roughly 23% to 37%).
  • The larger context notes that for proportions, standard MOE formulas often include a factor for the confidence level (e.g., 1.96 for 95% confidence) and the probability p(1−p); the transcript presents the simplified version for quick review.

Question 21: Sequences; first three terms (a_n)

  • Given the sequence with a first-term expression an = \frac{3}{n+1}, the first three terms are: [a1=\frac{3}{2}, \quad a2=\frac{3}{3}=1, \quad a3=\frac{3}{4}=\frac{3}{4}.]
  • Notation: the sequence terms are often written as a1, a2, a3, … with an representing the nth term.

Question 22: Series notation and explicit term form

  • The series is written with summation notation:
    [\sum{i=1}^{6} ai, \quad a_i = \text{the ith term}.
    ]
  • In the transcript, the explicit form for the nth term was derived as an = 3n+2 for a particular example, but the accompanying first-term values shown were consistent with the earlier an = 3/(n+1) example. The point is to identify the index, the range of summation, and the explicit formula for a_n so you can compute the sum efficiently.

Question 23: Sum of a short finite series

  • With terms from 2 through 5, you can evaluate term-by-term:
    [2^{4}-1=15,
    \3^{4}-1=80,
    \4^{4}-1=255,
    \5^{4}-1=624.
    ]
  • Sum = 15+80+255+624 = 974.
  • This illustrates that for a short finite number of terms, straightforward substitution and addition can be faster than applying a general formula.

Question 24: Explicit rule for an arithmetic sequence; find a_15

  • Given that the nth term of an arithmetic sequence has the form
    [an=a1+d(n-1).
    ] With the seventh term value a7=12 and common difference d=-3, solve for a1:
    [12=a1+6(-3)=a1-18 \Rightarrow a_1=30.
    ]
  • Therefore, the explicit rule is
    [a_n=30-3(n-1)=33-3n.
    ]
  • The fifteenth term is
    [a_{15}=33-3(15)=33-45=-12.]
    (There is a consistency check with the pattern: each term decreases by 3 from the previous term.)

Question 25: Arithmetic nth term given seventh term and difference

  • Given that the seventh term is 12 and the common difference is -3, the explicit rule is the same as above:
    [an=a1-3(n-1),] with a7=12 leading to a1=30 and hence
    [a_n=33-3n.]
  • This yields a_15 = 33-3(15) = -12 (as in Question 24).

Question 26: Geometric nth term; division by 3 (ratio 1/3)

  • A geometric sequence with first term a1 and common ratio r has [an=a1 r^{n-1}. ] The transcript notes a1=12 and r=1/3, so
    [a_n=12\left(\frac{1}{3}\right)^{n-1}.
    ]
  • For example, a8 would be [a8=12\left(\frac{1}{3}\right)^{7}.
    ]

Question 27: Geometric sequence from a given term and ratio

  • Given a third term a3=8 and ratio r=1/2, find a1:
    [a3=a1 r^{2}= a1 \left(\frac{1}{2}\right)^2 = \frac{a1}{4}=8 \Rightarrow a_1=32.
  • Therefore, the nth term is
    [a_n=32\left(\frac{1}{2}\right)^{n-1}.
    ]

Question 28: Sum of a finite geometric series

  • For a finite geometric series with first term $a1$ and common ratio $r$, the sum of the first $n$ terms is [Sn=\frac{a_1\left(1-r^{n}\right)}{1-r},\quad r\neq 1.]
  • Example from the transcript: $a1=3$, $r=2$, $n=10$ gives [S{10}=\frac{3\left(1-2^{10}\right)}{1-2}=\frac{3(1-1024)}{-1}=3069.]

Question 29: Infinite geometric series

  • When |r|<1, an infinite geometric series converges and the sum is
    [S=\frac{a_1}{1-r}.]
  • With a_1=100 and r=1/2, the sum is
    [S=\frac{100}{1-1/2}=\frac{100}{1/2}=200.]

Question 30: Evaluating trig functions on the unit circle (no calculator)

  • Key reference angles: 30°, 45°, 60° (or π/6, π/4, π/3).
  • Examples from the transcript:
    • sin(30°) = 1/2.
    • cos(5π/6) = -√3/2; therefore sin/cos/tan values follow from signs in the quadrant.
    • tan(-3π/4) = 1 (since tangent has period π and both sine and cosine are negative in the third quadrant, giving a positive ratio).
    • csc(2/10) and sec(7π/6) involve reciprocal relations; one notes the correct quadrant signs and then rationalizes denominators when needed.
    • cot(π) is undefined because tan(π)=0 (and cotangent is the reciprocal of tangent).
  • The unit circle is the primary tool for exact values without a calculator.

Question 31: Solve a right triangle (SOHCAHTOA)

  • Given a right triangle with a right angle, use sine, cosine, tangent relations:
    • If the opposite side is 2 and the hypotenuse is 5, then sin = 2/5. The missing side is (\sqrt{5^2-2^2}=\sqrt{25-4}=\sqrt{21}) (call this side adjacent).
    • Cosine: adjacent/hypotenuse = (\sqrt{21}/5).
    • Tangent: opposite/adjacent = 2/\sqrt{21} = \frac{2\sqrt{21}}{21}.
    • Cosecant, secant, and cotangent follow as reciprocals: csc = 5/2, sec = 5/\sqrt{21} = \frac{5\sqrt{21}}{21}, cot = \sqrt{21}/2.
  • The sine, cosine, and tangent values correspond to the given triangle; this is a standard SOHCAHTOA application on the unit circle / right-triangle context.

Question 32: Solve triangle ACT with Law of Sines and Law of Cosines

  • Given a non-right triangle with some side-angle data, use the Law of Sines and Law of Cosines to solve for all sides and angles. In the example:
    • One angle was found to be 70°, another angle found via the Law of Sines, and a side was found to be about 11.3 for the side opposite 62°. The other side, opposite 48°, came out about 9.5. The conclusion: all angles and sides can be determined using the Law of Sines and Law of Cosines depending on the given data.

Question 33: Angle of depression

  • The angle of depression is measured from the horizontal downward. Using a right-triangle model, if the opposite side is 1000 and the adjacent is 3000, then
    [\theta=\tan^{-1}\left(\frac{1000}{3000}\right)≈\tan^{-1}(0.333…)≈18.4^{\circ}.]

Question 34: Coterminal angles in radians

  • A positive angle coterminal with (\tfrac{\pi}{5}) can be obtained by adding multiples of 2\pi:
    [\tfrac{\pi}{5}+2\pi=\tfrac{11\pi}{5}.]
  • A negative coterminal angle can be obtained by subtracting multiples of 2\pi:
    [\tfrac{\pi}{5}-2\pi=\tfrac{\pi}{5}-\tfrac{10\pi}{5}= -\tfrac{9\pi}{5}.]
  • In degrees, add/subtract 360° to get a positive and a negative coterminal angle.

Question 35: Converting between radians and degrees

  • The key conversion is that 180^{\circ}=\pi\text{ radians}. To convert radians to degrees, multiply by \dfrac{180^{\circ}}{\pi}; to convert degrees to radians, multiply by \dfrac{\pi}{180^{\circ}}.
  • Example (radians to degrees): if you have \frac{7\pi}{12}\text{ rad}, degrees = \frac{7\pi}{12}\cdot\frac{180^{\circ}}{\pi}=\frac{7\cdot180}{12}=105^{\circ}.
  • Example (degrees to radians): if you have 105^{\circ}, radians = 105^{\circ}\cdot\frac{\pi}{180^{\circ}}=\frac{105\pi}{180}=\frac{7\pi}{12}.

Question 36: Arc length and area of a sector

  • Given a circle of radius R=4 and central angle (in radians) \theta=\frac{2\pi}{3}:
  • Arc length: s=R\theta=4\cdot\frac{2\pi}{3}= rac{8\pi}{3}.
  • Area of the sector: A=\frac{1}{2}R^{2}\theta=\frac{1}{2}(4)^2\cdot\frac{2\pi}{3}=\frac{16\pi}{3}.
  • An alternative method is to set up a proportion using the whole circle or to use the formulae directly if the angle is already in radians.

Question 37: Inverse trig values and principal values

  • Evaluate inverse trig with restricted domains:
    • (\sin^{-1}(-1) = -\frac{\pi}{2}) (in the range [-\frac{\pi}{2}, \frac{\pi}{2}]).
    • (\cos^{-1}(-\tfrac{\sqrt{3}}{2}) = \frac{5\pi}{6} (range [0,\pi]).
    • (\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} (range [-\frac{\pi}{2}, \frac{\pi}{2}]).

Question 38: Solve triangle ACT (Law of Sines/Cosines)

  • Given angle data and a side, use the Law of Sines to find an unknown side: e.g., if angle T is 62° and opposite side is a certain length, you compute other sides via
    [\frac{\sin 62^{\circ}}{\text{opposite}} = \frac{\sin 70^{\circ}}{12},] and solve for the unknown side. The transcript reports values like T ≈ 11.3 and A ≈ 9.5 as part of the solution process and emphasizes reviewing the Law of Sines and Law of Cosines for non-right triangles.

Question 41: Graphing sine with amplitude and period

  • General form: y=A\sin(Bx-C)+D. The transcript notes the amplitude A and the period P=\dfrac{2\pi}{|B|}. For the example: A=2, P=\dfrac{2\pi}{1/2}=4\pi. The left-right span of one period is from (-2\pi) to (2\pi) in terms of the horizontal scale, and the graph repeats every period. The domain is all real numbers, and the range is from (-|A|+D) to (|A|+D) (for the given example range is from (-2) to 2 around the midline). The note also emphasizes labeling the midline and using the phase shift if present.

Question 42: Graphing cosine with amplitude, period, phase shift, and vertical shift

  • For y=\tfrac{1}{2}\cos\left(\tfrac{1}{4}x+\tfrac{\pi}{4}\right)-1,
    • Amplitude |A|=\tfrac{1}{2}.
    • Period P=\dfrac{2\pi}{|B|}=\dfrac{2\pi}{1/4}=8\pi. The scale can be divided into quarters of the period (the transcript divides the period into eight segments with a step of \pi/8).
    • Phase shift is left by \frac{\pi}{4} (from the inside x+\frac{\pi}{4}).
    • Vertical shift is downward by 1 (the -1 outside the cosine).
  • The domain remains all real numbers; the range is shifted by the vertical amount, giving a vertical extent from -1-\tfrac{1}{2} to -1+\tfrac{1}{2}, i.e. from -\tfrac{3}{2} to -\tfrac{1}{2}, for the example.

Question 43: Graphing tangent; period and domain

  • Given y=2\tan\left(\tfrac{1}{2}x\right),
    • The tangent graph has vertical asymptotes where the argument equals \tfrac{\pi}{2}+k\pi, so the asymptotes occur at
      \tfrac{1}{2}x=\tfrac{\pi}{2}+k\pi \Rightarrow x=\pi+2k\pi.
    • Period is P=\dfrac{\pi}{|B|}=\dfrac{\pi}{1/2}=2\pi. The amplitude is not defined for tangent, but the graph is stretched by the factor of 2 in the y-direction.
    • Domain: all real numbers except the vertical asymptotes x=\pi+2k\pi\ (k\in\mathbb{Z}).
    • Range: all real numbers.
  • The transcript also provides a quick sketch strategy by using known points of tangent and adjusting for the factor of 2.

Question 44: Right-triangle trig given quadrant

  • Given cos(\theta)=\tfrac{3}{5} with (\theta) in Quadrant IV, draw a quick reference triangle: adjacent = 3, hypotenuse = 5, so opposite = (-\sqrt{5^2-3^2}) = -4).
  • Then sin(\theta) = opposite/hypotenuse = -\frac{4}{5}. (The unit-circle signs: cosine is positive in QIV, sine is negative.)

Question 45: Trig identities (simplification)

  • (a) \csc x\cdot\sin x=1; \sec x\cdot\cos x=1; hence \csc x\tan x=\frac{1}{\sin x}\cdot\frac{\sin x}{\cos x}=\frac{1}{\cos x}=\sec x. When simplified, the product of a trig function and its reciprocal cancels to 1 where defined.
  • (b) \sin^{2}x+\cos^{2}x=1, and since the reciprocal of sine is cosecant, the expression can be rewritten as \csc x.

Question 46: Solving in an interval; sine and tangent equations

  • Part A: Solve 2\sin^{2}x=1\quad(0\le x<2\pi). Then \sin^{2}x=\tfrac{1}{2} \Rightarrow \sin x=\pm\frac{\sqrt{2}}{2}. The solutions in \(0,2\pi) are x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}.
  • Part B: Solve 3\tan x-\sqrt{3}=0\Rightarrow \tan x=\frac{\sqrt{3}}{3}. The base angles corresponding to tangent value are x=\frac{\pi}{6} and x=\frac{7\pi}{6} (with period \pi yielding general solution x=\frac{\pi}{6}+k\pi, k\in\mathbb{Z}).
  • Part C (general solution): Factor the equation, set each factor to zero, and find all solutions in the interval, including extraneous solutions. The transcript notes that one factor leads to a value outside the sine range (extraneous), while the other factor gives valid solutions such as x=\frac{\pi}{6}+2\pi k\ \text{or}\ x=\frac{5\pi}{6}+2\pi k,\ k\in\mathbb{Z}.