Answer
$\theta=\{36.9^o,143.1^o,216.9^o,323.1^o\}$
Work Step by Step
$7sin^2(\theta)-9cos(2\theta)=0$
$7sin^2(\theta)-9[1-2sin^2(\theta)]=0$
$7sin^2(\theta)-9+18sin^2(\theta)=0$
$25sin^2(\theta)-9=0$
$sin^2(\theta)=\frac{9}{25}$
$sin(\theta)=\pm \frac{3}{5}$
$sin(\theta)=\frac{3}{5}$
$\theta=sin^{-1}(\frac{3}{5})$
We know $sin(\theta)$ is postive in quadrant $I$ and $II$
$\theta=36.9^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o-36.9^o=143.1^o$
$\theta=36.9^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=143.1^o$
$sin(\theta)=\frac{-3}{5}$
$\theta=sin^{-1}(\frac{-3}{5})$
We know $sin(\theta)$ is negative in quadrant $III$ and $IV$
$\theta=180^o+36.9^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-36.9^o=323.1^o$
$\theta=216.9^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=323.1^o$
$\theta=\{36.9^o,143.1^o,216.9^o,323.1^o\}$
