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.
Hint:
$${5\sin^2(x)\over\cos^2(x)}+{8\sin(x)\over\cos(x)}-{3\sin^2(x)+3\cos^2(x)\over \cos^2(x)}=0$$
$$2\tan^2(x)+8\tan(x)-3=0$$
Let $s:=\sin x$. The equation is
$$5s^2\pm8s\sqrt{1-s^2}-3=0$$
or after transformation,
$$64s^2(1-s^2)=9-30s^2+25s^4.$$
The solutions of the biquadratic equation are
$$s=\pm\sqrt{\frac{47\pm8\sqrt{22}}{89}}.$$