Question

Find all solutions in the set of real numbers. Show all your work. cos2θ=−sin2θ

Answer 1

`\begin{gathered} \text{Given} \\ \cos 2\theta=-\sin ^2\theta \end{gathered}`

Equate the given to zero

`\begin{gathered} \cos 2\theta=-\sin ^2\theta\Longrightarrow\cos 2\theta+\sin ^2\theta=0 \\ \\ \text{Use the identity }\cos 2\theta=1-2\sin ^2\theta,\text{ THEN the equation becomes} \\ \cos 2\theta+\sin ^2\theta=0 \\ (1-2\sin ^2\theta)+\sin ^2\theta=0 \\ \\ \text{simplify} \\ (1-2\sin ^2\theta)+\sin ^2\theta=0 \\ 1-2\sin ^2\theta+\sin ^2\theta=0 \\ 1-\sin ^2\theta=0 \\ \\ \text{Add }\sin ^2\theta\text{ to both sides} \\ 1-\sin ^2\theta+\sin ^2\theta=0+\sin ^2\theta \\ 1\cancel{-\sin ^2\theta+\sin ^2\theta}=\sin ^2\theta \\ 1=\sin ^2\theta \\ \text{OR} \\ \sin ^2\theta=1 \\ \\ \text{get the square root of both sides} \\ \sqrt[]{\sin ^2\theta}=\sqrt[]{1} \\ \sin \theta=\pm1 \\ \\ \text{The values for which }\sin \theta\text{ is equal to }+1\text{ or }-1\text{ is} \\ \theta=(\pi)/(2)+2\pi n,\theta=(3\pi)/(2)+2\pi n \end{gathered}`