How do you solve $5\sin^2(x)+8\sin x \cos x-3=0$? I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into $\sin 2x$ but then I have $\sin^2(x)$ and $\sin 2 x$ together.

Rewrite it as $5\sin^2(x)+8\sin x \cos x-3(\sin^2 x + \cos^2 x)$, which gives $2 \sin^2 x + 8 \sin x \cos x - 3 \cos^2 x$, a quadratic in $\sin x$.
Using the quadratic formula, the roots of $2u^2 + 8u - 3$ are $-2 - \sqrt{11/2}$ and $-2 + \sqrt{11/2}$. Hence this is equivalent to $(u + 2 + \sqrt{11/2})(u + 2 - \sqrt{11/2}) = 0$, so the original factorises as $(\sin x + (2 + \sqrt{11/2})\cos x)(\sin x + (2 - \sqrt{11/2})\cos x) = 0$.
