Question

Решите уравнение
1) 2sin²x sin x= 0
2)4sinx²- 8sinx·cosx+10cos²x= 3

Answer 1

$1)\;2\sin^2x-\sin x=0\\\sin x(2\sin x-1)=0\\\begin{cases}\sin x=0\\2\sin x-1=0\end{cases}\Rightarrow\begin{cases}\sin x=0\\\sin x=\frac12\end{cases}\Rightarrow\begin{cases}x=\pi n\\x=(-1)^n\cdot\frac\pi6+\pi n\end{cases},\;n\in\mathbb{Z}$

$2)\;4\sin^2x-8\sin x\cos x+10\cos^2x=3\\4(\sin^2x-2\sin x\cos x+\cos^2x)+6\cos^2x-3=0\\4(\sin x-\cos x)^2+6\cos^2x-3\sin^2x-3\cos^2x=0\\4(\cos x-\sin x)^2+3(\cos^2x-\sin^2x)=0\\4(\cos x-\sin x)^2+3(\cos x-\sin x)(\cos x+\sin x)=0\\(\cos x-\sin x)(4(\cos x-\sin x)+3(\cos x+\sin x))=0\\(\cos x-\sin x)(4\cos x-4\sin x+3\cos x+3\sin x)=0\\(\cos x-\sin x)(7\cos x-\sin x)=0$
$\begin{cases}\cos x-\sin x=0\\7\cos x-\sin x=0\end{cases}\Rightarrow\begin{cases}tg x=1\\tgx=7\end{cases}\Rightarrow\begin{cases}x=\frac\pi4+\pi k\\x=arctg(7)+\pi k\end{cases},k\in\mathbb{Z}$