Question

16. 0.01 points l Previous Answers SPreCalc77.T017. Solve the trigonometric equation in the interval [0, 2π). Give the exact value, if possible; otherwise, round your answer to two de sin(20)-cos(θ) = 0 Submit Answer Save Progress 17、0 1/1 points ! Previous Answers SPreCalc72.T014

Answer 1

sin(2$\theta$) - cos($\theta$) = 0

=> 2sin($\theta$)cos($\theta$) - cos($\theta$) = 0

=> cos($\theta$) (2sin($\theta$) - 1) = 0

=> cos($\theta$) = 0 or (2sin($\theta$) - 1) = 0

cos($\theta$) = 0

We know that cos x = 0 for x = $\pi$/2 and 3$\pi$/2

Therefore, solutions of $\theta$ for which cos($\theta$) = 0 is given by

$\theta$ = $\pi$/2, $\theta$ = 3$\pi$/2

(2sin($\theta$) - 1) = 0

=> sin $\theta$ = 1/2

We know that sine function is positive in the first and second quadrant and sin x = 1/2 for x = $\pi$/6

Reference angle in the first quadrant = x

Reference angle in the second quadrant = $\pi$ - x

Therefore, solutions of $\theta$ for which (2sin($\theta$) - 1) = 0 is given by

$\theta$ = $\pi$/6

and, $\theta$ = $\pi$ - $\pi$/6 = 5$\pi$/6

Combining all the solutions of $\theta$ we get

$\theta$ = $\pi$/6, $\pi$/2, 5$\pi$/6, 3$\pi$/2

Note : The solution for the equation is to be provided in the interval [0, 2$\pi$)