Question

Sin^2x-3sin x cos x+2cos^2x=0

Answer 1

$\sin^2x-3\sin x\cos x+2\cos^2x =0\\(\cos x-\sin x)(2\cos x - \sin x) = 0\\\left[\begin{array}{ccc}\cos x = \sin x\\2\cos x = \sin x\end{array}\right.\\\left[\begin{array}{ccc}\b{tg}~x=1\\\b{tg}~ x = 2\end{array}\right.\\\left[\begin{array}{ccc}x=\dfrac\pi4+\pi n\\x=\b{arctg}~2+\pi n\end{array}\right.,n\in\mathbb Z\\$

$\left[\begin{array}{ccc}x=\dfrac\pi4+\pi n\\x=\b{arctg}~2+\pi n\end{array}\right.,n\in\mathbb Z$