# Math Help - trigonometric equation

1. ## trigonometric equation

solve for x
$4(\cos^3 x + \sin^3 x) = \cos x + 3 \sin x$

2. It is well known that $\sin^2x+\cos^2x=1$. Using this the equation can be written as

$4(\cos^3x+\sin^3x)=\cos x(\sin^2x+\cos^2x)+3\sin x(\sin^2x+\cos^2x)$

$\sin^3x-\sin^2x\cos x-3\sin x\cos^2x+3\cos^2x=0$

$\sin^2x(\sin x-\cos x)-3\cos^2x(\sin x-\cos x)=0$

$(\sin x-\cos x)(\sin^2x-3\cos^2x)=0$

Now, $\sin x-\cos x=0\Rightarrow\tan x=1\Rightarrow x=\frac{\pi}{4}+k\pi, \ k\in\mathbb{Z}$

or $\sin^2x-3\cos^2x=0\Rightarrow\tan^2x=3\Rightarrow \tan x=\pm\sqrt{3}\Rightarrow x=\pm\frac{\pi}{3}+k\pi, \ k\in\mathbb{Z}$

3. good, thanks red dog much.