2;3;4буду благодарен

Решите задачу:

$2^o\\2\cos^23\alpha tg3\alpha=5\cos^23\alpha\cdot\frac{\sin3\alpha}{\cos\3\alpha}=5\cos3\alpha\sin3\alpha=2,5\cdot\sin6\alpha\\\\3.\\\sin^2\alpha+\cos^\alpha=1\Rightarrow\cos\alpha=\sqrt{1-\sin^2\alpha}\\\cos\alpha=\sqrt{1-0,64}=\sqrt{0,36}=\pm0,6\\\frac\pi2<\alpha\pi\Rightarrow\cos\alpha<0\\\cos\alpha=-0,6$

$4.\\\frac{\sin(x+45^o)+\sin(x-45^o)}{\sin(x+45^o)-\sin(x-45^o)}=\frac{\sin x\cos45^o+\cos x\sin45^o+\sin x\cos45^o-\cos x\sin45^o}{\sin x\cos45^o+\cos x\sin45^o-\sin x\cos45^o+\cos x\sin45^o}=\\=\frac{2\sin x\cos45^o}{2\cos x\sin45^o}=\frac{\sqrt2\sin x}{\sqrt2\cos x}=tgx$