# Math Help - Where's the mistake?

1. ## Where's the mistake?

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

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

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

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

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

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

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

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

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

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

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

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

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

Thanks for discovering the mistake.

2. 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.

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

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

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

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

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

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

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

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

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

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

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

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

Thanks for discovering the mistake.
When squaring both sides of an equation extraneous solutions can be introduced. The solutions you got therefore need to be checked by substitution in the original equation. You will find some satisfy it and some don't .......

3. $\frac{\pi}{6}$ an extraneous solution ...

$\sin\left(\frac{\pi}{6}\right) + \sqrt{3}\cos\left(\frac{\pi}{6}\right) = \frac{1}{2} + \frac{3}{2} \neq 1$