Trig solving Review

Trig solving Review

10th Grade - University

46 Qs

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Trig solving Review

Trig solving Review

Assessment

Quiz

Mathematics

10th Grade - University

Medium

CCSS
HSF.TF.B.7

Standards-aligned

Created by

Rachael Kowalski

Used 3+ times

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46 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Which of the following is NOT a solution to
 csc θ=23\csc\ \theta=\frac{2}{\sqrt{3}} on  [0,4π)\left[0,4\pi\right)   ?

 π3\frac{\pi}{3}  

 2π3\frac{2\pi}{3}  

 5π3\frac{5\pi}{3}  

 7π3\frac{7\pi}{3}  

2.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Solve sinθ=12\sin\theta=\frac{1}{2}   over the interval [0, 2π)

 θ=π6,7π6\theta=\frac{\pi}{6},\frac{7\pi}{6}  

 θ=π6,5π6\theta=\frac{\pi}{6},\frac{5\pi}{6}  

 θ=5π6,7π6\theta=\frac{5\pi}{6},\frac{7\pi}{6}  

 θ=7π6,11π6\theta=\frac{7\pi}{6},\frac{11\pi}{6}  

3.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

 Solve 3sec(θ) = 63\sec\left(\theta\right)\ =\ -6 over the interval [0, 2π)\left[0,\ 2\pi\right)  

 θ =π3, 2π3\theta\ =\frac{\pi}{3},\ \frac{2\pi}{3}  

 θ=π3,5π3\theta=\frac{\pi}{3},\frac{5\pi}{3}  

 θ=2π3,4π3\theta=\frac{2\pi}{3},\frac{4\pi}{3}  

 θ=4π3,5π3\theta=\frac{4\pi}{3},\frac{5\pi}{3}  

4.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Solve  4sin(x)+2 = 2sin(x)4\sin\left(x\right)+2\ =\ 2\sin\left(x\right)  over  [0, 2π)\left[0,\ 2\pi\right)  

 x=π2, 3π2x=\frac{\pi}{2},\ \frac{3\pi}{2}  

 x=π2x=\frac{\pi}{2}  

 x=3π2x=\frac{3\pi}{2}  

 x=πx=\pi  

5.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Which of these is NOT a solution to tan θ = 0 ?

θ=0\theta=0

θ=π2\theta=\frac{\pi}{2}

θ=π\theta=\pi

θ=2π\theta=2\pi

6.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Find the complete solution for  sec2θ=2\sec^2\theta=2  over [0, 2π)

 θ=π4, 7π4\theta=\frac{\pi}{4},\ \frac{7\pi}{4}  

 θ=π4, 3π4\theta=\frac{\pi}{4},\ \frac{3\pi}{4}  

 θ=3π4,5π4\theta=\frac{3\pi}{4},\frac{5\pi}{4}  

 θ=π4,3π4,5π4,7π4\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}  

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Solve  3cotθ+3 =0\sqrt{3}\cot\theta+3\ =0 over (, )\left(-\infty,\ \infty\right)   

 θ=π3+πk\theta=\frac{\pi}{3}+\pi k 

 θ=2π3+πk\theta=\frac{2\pi}{3}+\pi k  

 θ=π6+πk\theta=\frac{\pi}{6}+\pi k  

 θ=5π6+πk\theta=\frac{5\pi}{6}+\pi k  

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