Answer
$\theta=\{34.4^o,145.6^o,214.4^o,325.6^o\}$
Work Step by Step
$16cos(2\theta)-18sin^2(\theta)=0$
$16[1-2sin^2(\theta)]-18sin^2(\theta)=0$
$16-32sin^2(\theta)-18sin^2(\theta)=0$
$-50sin^2(\theta)+16=0$
$sin^2(\theta)=\frac{-16}{-50}$
$sin(\theta)=\pm \frac{4}{5\sqrt{2}}$
$sin(\theta)=\frac{4}{5\sqrt{2}}$
$\theta=sin^{-1}(\frac{4}{5\sqrt{2}})$
We know $sin(\theta)$ is positive in quadrant $I$ and $II$
$\theta=34.4^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o-34.4^o=145.6^o$
$\theta=34.4^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=145.6^o$
$sin(\theta)=\frac{-4}{5\sqrt{2}}$
$\theta=sin^{-1}(\frac{-4}{5\sqrt{2}})$
We know $sin(\theta)$ is negative in quadrant $III$ and $IV$
$\theta=180^o+34.4^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-34.4^o=325.6^o$
$\theta=214.4^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=325.6^o$
$\theta=\{34.4^o,145.6^o,214.4^o,325.6^o\}$
