Помогите решить: 1) 2sin^2x-sinx-2=0 2) sin2x-cosx=0 3) cos7x+cosx=0 4) tgx-2ctgx=0

1\\\sin x=\dfrac{1-\sqrt{17}}{4}\Rightarrow \boxed{x=(-1)^k\arcsin\frac{1-\sqrt{17}}{4}+\pi k, \ k\in Z}" alt="2\sin^2x-\sin x-2=0\\D=(-1)^2-4\cdot2\cdot(-2)=17\\\sin x\neq \dfrac{1+\sqrt{17}}{4}>1\\\sin x=\dfrac{1-\sqrt{17}}{4}\Rightarrow \boxed{x=(-1)^k\arcsin\frac{1-\sqrt{17}}{4}+\pi k, \ k\in Z}" align="absmiddle" class="latex-formula">

$\sin2x-\cos x=0\\2\sin x\cos x-\cos x=0\\\cos x(2\sin x-1)=0\\\cos x=0\Rightarrow \boxed{x_1=\dfrac{\pi}{2} +\pi n, \ n\in Z}\\2\sin x-1=0\Rightarrow \sin x=\dfrac{1}{2} \Rightarrow \boxed{x_2=(-1)^k\dfrac{\pi}{6} +\pi k, \ k\in Z}\\$

$\cos7x+\cos x=0\\\\2\cos\dfrac{7x+x}{2}\cos\dfrac{7x-x}{2}=0\\\\2\cos4x\cos3x=0\\\cos4x=0\Rightarrow 4x=\dfrac{\pi}{2} +\pi n \Rightarrow \boxed{x_1=\dfrac{\pi}{8} +\frac{\pi n}{4}, \ n\in Z}\\\cos3x=0\Rightarrow 3x=\dfrac{\pi}{2} +\pi n \Rightarrow \boxed{x_2=\dfrac{\pi}{6} +\frac{\pi n}{3}, \ n\in Z }$

$\mathrm{tg}x-2\mathrm{ctg}x=0 \ (\sin x\neq 0; \cos x\neq 0) \\\mathrm{tg}x-\dfrac{2}{\mathrm{tg}x}=0\\\mathrm{tg}^2x-2=0\\\mathrm{tg}^2x=2\\\mathrm{tg}x=\pm\sqrt{2} \\\boxed{x=\pm\mathrm{arctg}\sqrt{2}+\pi n, \ n\in Z}$