Question

Решите уравнение:
3sin^3x + 4sin^2x cos x - sinx cos^2x = 2sinx + 3cos x

Answer 1

$3\sin^3x + 4\sin^2x \cos x - \sin x \cos^2x = 2\sin x + 3\cos x;\\3\sin^3x + 4\sin^2x \cos x - \sin x \cos^2x = 2(\sin^2x+\cos^2x)\sin x + 3(\sin^2x+\cos^2x)\cos x;\\\sin^3x + \sin^2x \cos x - 3\sin x \cos^2x -3\cos^3x=0;\\\sin^2x(\sin x+\cos x)-3\cos^2x(\sin x+\cos x)=0;\\(1-\cos^2 x)(\sin x+\cos x)-3\cos^2x(\sin x+\cos x)=0;\\(1-4\cos^2 x)(\sin x+\cos x)=0;$

$\left[\begin{matrix}\cos x=\pm \frac{1}{2},\\ \sin (x+\frac{\pi}{4})=0;\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\pm \frac{\pi}{3}+2\pi n,\,n\in Z,\\x=\pm \frac{2\pi}{3}+2\pi m,\,m\in Z,\\ x=-\frac{\pi}{4}+\pi k,\,k\in Z.\end{matrix}\right.\Rightarrow$