√3 sin 2x .(2cos x +1)+2= cos 3x + cos 2x -3cos x 04/08/2021 Bởi Skylar √3 sin 2x .(2cos x +1)+2= cos 3x + cos 2x -3cos x
Giải thích các bước giải: $\sqrt{3}\sin 2x(2\cos x+1)+2=\cos 3x+\cos 2x-3\cos x$ $\to \sqrt{3}\sin 2x(2\cos x+1)+2=4\cos^3x-3\cos x+2\cos^2x-1-3\cos x$ $\to \sqrt{3}\sin 2x(2\cos x+1)+2=4\cos^3x+2\cos^2x-6\cos x-1$ $\to \sqrt{3}\sin 2x(2\cos x+1)=4\cos^3x+2\cos^2x-6\cos x-3$ $\to \sqrt{3}\sin 2x(2\cos x+1)=(2\cos x+1)(2\cos^2x-3)$ $\to 2\cos x+1=0\to x\in\{\dfrac{2\pi}{3}+2k\pi,\dfrac{4\pi}{3}+2k\pi \}$ Hoặc $\sqrt{3}\sin 2x=2\cos^2x-3$ $\to \sqrt{3}\sin 2x=2\cos^2x-3(\sin^2x+\cos^2x)$ $\to 2\sqrt{3}\sin x\cos x=-\cos^2x-3\sin^2x$ $\to \cos^2x+2\sqrt{3}\sin x\cos x+3\sin^2x=0$ $\to (\cos x+\sqrt{3}\sin x)^2=0$ $\to \cos x+\sqrt{3}\sin x=0$ $\to \dfrac{\sin x}{\cos x}=-\dfrac{1}{\sqrt{3}}$ $\to \tan x=-\dfrac{1}{\sqrt{3}}$ $\to x=\dfrac{5\pi}{6}+k\pi$ Bình luận
Giải thích các bước giải:
$\sqrt{3}\sin 2x(2\cos x+1)+2=\cos 3x+\cos 2x-3\cos x$
$\to \sqrt{3}\sin 2x(2\cos x+1)+2=4\cos^3x-3\cos x+2\cos^2x-1-3\cos x$
$\to \sqrt{3}\sin 2x(2\cos x+1)+2=4\cos^3x+2\cos^2x-6\cos x-1$
$\to \sqrt{3}\sin 2x(2\cos x+1)=4\cos^3x+2\cos^2x-6\cos x-3$
$\to \sqrt{3}\sin 2x(2\cos x+1)=(2\cos x+1)(2\cos^2x-3)$
$\to 2\cos x+1=0\to x\in\{\dfrac{2\pi}{3}+2k\pi,\dfrac{4\pi}{3}+2k\pi \}$
Hoặc $\sqrt{3}\sin 2x=2\cos^2x-3$
$\to \sqrt{3}\sin 2x=2\cos^2x-3(\sin^2x+\cos^2x)$
$\to 2\sqrt{3}\sin x\cos x=-\cos^2x-3\sin^2x$
$\to \cos^2x+2\sqrt{3}\sin x\cos x+3\sin^2x=0$
$\to (\cos x+\sqrt{3}\sin x)^2=0$
$\to \cos x+\sqrt{3}\sin x=0$
$\to \dfrac{\sin x}{\cos x}=-\dfrac{1}{\sqrt{3}}$
$\to \tan x=-\dfrac{1}{\sqrt{3}}$
$\to x=\dfrac{5\pi}{6}+k\pi$