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1 -1 ≤ x ≤ 1
2 all reals
3 x ≤ -1 or x ≥ 1
4 0° ≤ y ≤ 180°
5 -90° < y < 90°
1. domain of arcsinxx
2. domain of arccotx
3. domain of arccscx
4. range of arccosxall reals
5. range of arctanx
https://cdn.lti.glynlyon.com/media/07e914b4-1ded-44f3-b133-16e33b06f543/img/graph_arctanx.gif
Click on the graph below to choose the graph of the function y = arctan(x).
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2 is not in the domain of arccosine.
Simplify arccos2.
-60°
240° is not in the range of arcsine, so find a coterminal angle that is.
Simplify arcsin(sin240°).
π/3
The angle whose secant is 2 has a cosine of 1/2.
Simplify arcsec2.
60
Simplify arccot √3/3°
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240° is not in the range of arcsine, so find a coterminal angle that is.
Simplify sec(arcsec1/2).
168.5°
csc 2θ - cscθ - 20 = 0
(cscθ - 5)(cscθ + 4) = 0
cscθ = 5 or cscθ = -4 (sinθ = 1/5 or sinθ = -1/4)
Use a calculator in degree mode:
5 1/x INV sin
This gives 11.5°.
Sine is also positive in Quadrant II:
180 - 11.5 = 168.5°
Use a calculator in degree mode:
4 +/- 1/x INV sin
This gives -14.5°, or 360 - 14.5 = 345.5°.
Sine is also negative in Quadrant III:
180 + 14.5 = 194.5°
Which of the following angles is in the solution set of csc2θ - cscθ - 20 = 0 for 0° ≤ θ < 360°.
15/17
The range of arctangent is -90° to 90°, so arctan(-8/15) results in a fourth-quadrant angle.
Tangent is opposite over adjacent.
Find the hypotenuse:8^2 + 15^2 = c^2
c = 17
Since the angle is a fourth-quadrant angle, the cosine is positive. Cosine is adjacent over hypotenuse: 15/17.
Evaluate cos[arctan(-8/15)].
4
sec 2
θ = 2secθ = ± √2
√2 is in the domain of secant, so there is an angle in each quadrant.
Find the number of solutions for the given equation for 0° ≤ θ ≤ 360°.
sec2θ = 2
-π/4
sec π = 1/cos π = 1/(-1) = -1
The range of arctangent is -π/2 to π/2, and the angle whose tangent is 1 is π/4.
arctan(sec π) = arctan(-1) = -π/4
Evaluate arctan(secπ).
{5π/6, 11π/6}
3tan x +√3 = 0
3tan x = -√3
tan x = - √3/3
The reference angle is π/6, and tangent is negative in Quadrants II and IV.
Solve 3tanx + √3 = 0 for 0 ≤ x < 2π.
300
180
60
secθsinθ - 2sinθ = 0
sinθ(secθ - 2) = 0
sinθ = 0 or secθ = 2 (implies cosθ = 1/2)
θ = 0, 180° or θ = 60°, 300°
Select all the solutions of the given equation for 0° < θ < 360°.
8
1/4 cot x - 11 = -9
1/4cot x = 2
cot x = 8
If 1/4 cotx - 11 = -9, then cotx = _____.
{77°}
https://assets.learnosity.com/organisations/328/ea9e9c34-818c-4d34-a124-ac58e2c9b6b8/soln_2cossqx_plus4cosx_minus1.gif
Use a calculator in degree mode:
2 +/- + 6 INV x^2 = ÷ 2 = INV cos
This gives 77.0°.
The other solution is not in the domain of arccosine.
Solve 2cos2x + 4cosx - 1 = 0 for 0° ≤ x < 180°.
