Можете помочь))) с1; с2 и с3. Очень надо

Решите задачу:

$C1.\;4\cos x\sin x-3\sin^2x=1\\ 4\cos x\sin x-3\sin^2x=\cos^2x+\sin^2x\\ \cos^2x+\sin^2x-4\cos x\sin x+3\sin^2x=0\\ \cos^2x-4\cos x\sin x+4\sin^2x=0\\ (\cos x-2\sin x)^2=0\\ \cos x-2\sin x=0\\ 2\sin x=\cos x\\ tgx=\frac12\\ x=arctg\frac12\approx27^o=\frac{3\pi}{20}+\pi n,\;n\in\mathbb{Z}$

$x\in\left[-\frac{3\pi}2;\;\frac{5\pi}4\right]\Rightarrow -\frac{3\pi}2\leq\frac{3\pi}{20}+\pi n\leq\frac{5\pi}4\\ -\frac{3\pi}{2}-\frac{3\pi}{20}\leq\pi n\leq\frac{5\pi}4-\frac{3\pi}{20}\\ -\frac{33\pi}{20}\leq\pi n\leq\frac{22\pi}{20}\\ -\frac{33}{20}\leq n\leq\frac{22}{20}\\ -1,65\leq n\leq 1,1\\ n\in\mathbb{Z}\Rightarrow n\in[-1;1]:\\ n=-1\Rightarrow x = \frac{3\pi}{20}-\pi=-\frac{17\pi}{20}\\ n=0\Rightarrow x=\frac{3\pi}{20}\\ n=1\Rightarrow x = \frac{3\pi}{20}+\pi=\frac{23\pi}{20}$