Trigonometry! #137

$\sec^{2}\theta+\csc^{2}\theta=8 \\ \dfrac{1}{\cos^{2}\theta}+\dfrac{1}{\sin^{2}\theta}=8 \\ \dfrac{\sin^{2}\theta+\cos^{2}\theta}{\sin^{2}\theta\cos^{2}\theta}=8 \\ 8\sin^{2}\theta\cos^{2}\theta=1 \\ 2(2\sin\theta\cos\theta)^{2}=1 \\ sin^{2}2\theta=\dfrac{1}{2} \\ \sin2\theta=\pm\dfrac{1}{\sqrt{2}}$

For $\sin2\theta=\dfrac{1}{\sqrt{2}}$ ,

$2\theta=(-1)^{n}\dfrac{\pi}{4}+n\pi \\ \theta=(-1)^{n}\dfrac{\pi}{8}+\dfrac{n\pi}{2}$

which gives three roots.

For $\sin2\theta=-\dfrac{1}{\sqrt{2}}$

$2\theta=(-1)^{n}\left(-\dfrac{\pi}{4}\right)+n\pi \\ \theta=(-1)^{n+1}\dfrac{\pi}{8}+\dfrac{n\pi}{2}$

which gives one root.

Hence we have $\boxed{four}$ roots.