Originally Posted by

**achacy** I 've solved the problem below but its incorrect. Could yo help me to find the mistake please, cause I can't see it.

$\displaystyle sinx + \sqrt{3}cosx=1 $

$\displaystyle sin x=1- \sqrt{3} cosx$

$\displaystyle sin ^{2} x+cos ^{2}x=1 $

$\displaystyle \left(1- \sqrt{3}cos x \right) ^{2}+cos ^{2} x=1 $

$\displaystyle 1+3cos ^{2}x-2 \sqrt{3}cos x +cos ^{2}x-1=0$

$\displaystyle 4cos ^{2} x-2 \sqrt{3}cos x=0$

$\displaystyle cos x \left(4cos x-2 \sqrt{3} \right)=0 $

$\displaystyle cos x =0 \vee 4cos x -2 \sqrt{3}=0$

$\displaystyle cos x =0 \vee cos x = \frac{ \sqrt{3} }{2} $

$\displaystyle x= \frac{\Pi}{2}+k\Pi \vee x= \frac{\Pi}{6}+2k\Pi \vee x=- \frac{\Pi}{6}+2k\Pi $

That's what I've got. However the solution should be as follows:

$\displaystyle x=- \frac{\Pi}{6}+2k\Pi \vee x= \frac{\Pi}{2}+2k\Pi $

Thanks for discovering the mistake.