Помогите с тригонометр .уровн.

1. в) 2. б) [latex]3.\;\sin x=\frac{\sqrt3}2\\x=(-1)^n\cdot\frac\pi3+\pi n,\;n\in\mathbb{Z}\\\\4.\;\cos x=-\frac{\sqrt2}2\\x=\pm\frac{3\pi}4+2\pi n,\;n\in\mathbb{Z}\\\\5.\;tg\left(2x+\frac\pi3\right)=1\\2x+\frac\pi3=\frac\pi4+\pi n\\2x=-\frac\pi{12}+\pi n\\x=-\frac\pi{24}+\frac{\pi n}2,\;n\in\mathbb{Z}\\\\6.\;\cos^2x-4\cos x+3=0\\\cos x=t,\;\cos^2x=t^2,\;-1\leq t\leq1\\t^2-4t+3=0\\D=16-4\cdot3=4\\t_{1,2}=\frac{4\pm2}2\\t_1=1\\t_1=3\;-\;He\;nogx.\\\cos x=1\\x=1\pi n,\;n\in\mathbb{Z}[/latex]. [latex]7.\;f(x)=2\sin\left(\frac\pi6-x\right)-\sqrt2\\2\sin\left(\frac\pi6-x\right)-\sqrt2=0\\\sin\left(\frac\pi6-x\right)=\frac{\sqrt2}2\\\frac\pi6-x=(-1)^n\cdot\frac\pi4+\pi n\\x=\frac\pi 6-(-1)^n\cdot\frac\pi4-\pi n,\;n\in\mathbb{Z}\\\\8.\;3\cos^2x+2\sin x\cos x=\sin^2x\\3+2tgx=tg^2x\\tx^2x-2tgx-3=0\\tgx=t,\;tg^2x=t^2\\t^2-2t-3=0\\D=4+4\cdot3=16\\t_{1,2}=\frac{2\pm4}2\\t_1=-1,\;t_2=3\\x_1=\frac{3\pi}4+\pi n\\x_2=arctg(3)+\pi n,\;n\in\mathbb{Z}[/latex] [latex]9.\;\left|\sin\left(2x+\frac\pi{10}\right)-\frac12\right|=1\\\begin{cases}\sin\left(2x+\frac\pi{10}\right)-\frac12=1\\\sin\left(2x+\frac\pi{10}\right)-\frac12=-1\end{cases}\Rightarrow\begin{cases}\sin\left(2x+\frac\pi{10}\right)=\frac32\ \textgreater \ 1\;-\;pew.HET.\\\sin\left(2x+\frac\pi{10}\right)=-\frac12\end{cases}\\\sin\left(2x+\frac\pi{10}\right)=-\frac12\\2x+\frac\pi{10}=(-1)^n\cdot\frac{7\pi}6+\pi n\\2x=(-1)^n\cdot\frac{7\pi}6-\frac\pi{10}+\pi n\\x=(-1)^n\cdot\frac{7\pi}{12}-\frac\pi{20}+\frac{\pi n}2[/latex]. [latex]10.\;2\cos(25^o-2x)+\sqrt3=0\\\cos(2x-25^o)=-\frac{\sqrt3}2\\2x-25^o=\pm\frac{5\pi}6+2\pi n\\2x=\pm\frac{5\pi}6+25^o+2\pi n\\x=\pm\frac{5\pi}{12}+12,5^o+\pi n\\\frac{5\pi}{12}=75^o\\n=0:\\x=-75^o+12,5^o=-62,5^o[/latex]