Question

Solve the equation on the interval $0 \le \theta < 2\pi$ $\cos(2\theta) + 8\sin^2\theta = 4$ What is the solution in the interval $0 < \theta < 2\pi$? Select the correct choice and fill in any answer boxes in your choice below A. $\theta = $ (Simplify your answer. Type an exact answer, using $\pi$ as needed. Type your answer in radians. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. There is no solution

          Solve the equation on the interval $0 \le \theta < 2\pi$
$\cos(2\theta) + 8\sin^2\theta = 4$
What is the solution in the interval $0 < \theta < 2\pi$? Select the correct choice and fill in any answer boxes in your choice below
A. $\theta = $
(Simplify your answer. Type an exact answer, using $\pi$ as needed. Type your answer in radians. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)
B. There is no solution

$Solve the equation on the interval 0 ≤θ < 2π cos(2θ) + 8sin^2θ = 4 What is the solution in the interval 0 < θ < 2π? Select the correct choice and fill in any answer boxes in your choice below A. θ = (Simplify your answer. Type an exact answer, using π as needed. Type your answer in radians. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. There is no solution$

Added by Lisa V.

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