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  5. 5.5 Trigonometric Equations
5.5 Trigonometric Equations

5.5 Trigonometric Equations

Assessment

Presentation

•

Mathematics

•

11th Grade

•

Easy

Created by

Larry Cooper

Used 1+ times

FREE Resource

9 Slides • 11 Questions

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​5.5 Trigonometric Equations
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Multiple Choice

The value of  sin⁡−1(−12) \sin^{-1}\left(-\frac{1}{2}\right)\  #39

1

π3\frac{\pi}{3}  

2

π6\frac{\pi}{6}  

3

−π3-\frac{\pi}{3}  

4

−π6-\frac{\pi}{6}  

7

Multiple Choice

Solve sinθ = ½ for all Radian angles between 0 and 2π2\pi  . #1

1

θ = π / 6,  7π / 6

2

θ = π / 6,  5π / 6

3

θ = 5π / 6,  7π / 6

4

θ = 7π / 6,  11π / 6

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Multiple Choice

Solve sinθ = -1 for all Radian angles between 0 and 2π2\pi  . #2

1

θ = π /2, 3π /2 

2

θ = π /2

3

θ = 3π /2

4

θ = π  

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Multiple Select

Solve for 0≤x≤2π

sinxtanx = sinx #7

1

π/2, 3π/2

2

3π/4, 7π/4

3

0, π, 2π

4

π/4, 5π/4

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Multiple Choice

Solve for 0≤x≤2π

2sin2x - 5sinx + 2 = 0 #8

1

π/6, 5π/6

2

7π/6, 11π/6

3

π/3, 2π/3

4

4π/3, 5π/3

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Multiple Select

Solve for 0≤x≤2π

2sin2x - sinx = 0 #9

1

0, π, 2π

2

π/2, 3π/2

3

π/6, 5π/6

4

π/3, 2π/3

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Multiple Choice

Solve for 0≤x≤2π

3sin2x + sinx - 4 = 0 #11

1

0, π

2

π/2

3

π/6, 5π/6

4

3π/2

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Multiple Select

Solve for 0≤x≤2π


2cos2x + 3cosx = -1

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0, 2π

2

π

3

2π/3, 4π/3

4

π/3, 5π/3

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Multiple Choice

Solve for 0≤x≤2π

tanx = 1 #10

1

π/2, π

2

π/4, 3π/4

3

π/4, 5π/4

4

3π/4, 7π/4

19

Multiple Select

Which of the following are the correct solutions for the trigonometric equation in the interval  (0, 2π]\left(0,\ 2\pi\right] ? 

CHOOSE ALL THAT APPLY (2sin⁡2x−1)(3tan⁡2x−1)=0\left(2\sin^2x-1\right)\left(3\tan^2x-1\right)=0  #18

1

π4,3π4\frac{\pi}{4},\frac{3\pi}{4}  

2

π6,5π6\frac{\pi}{6},\frac{5\pi}{6}  

3

7π6,11π6\frac{7\pi}{6},\frac{11\pi}{6}  

4

5π4,7π4\frac{5\pi}{4},\frac{7\pi}{4}  

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Multiple Choice

Give a general formula for all solutions to the following equation: 2cos⁡x+1=12\cos x+1=1  #33

1

π2+2πk, 3π2+πk\frac{\pi}{2}+2\pi k,\ \frac{3\pi}{2}+\pi k  

2

π2+πk\frac{\pi}{2}+\pi k  

3

π3+2πk, 4π3+2πk\frac{\pi}{3}+2\pi k,\ \frac{4\pi}{3}+2\pi k  

4

3π2+2πk\frac{3\pi}{2}+2\pi k  

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