# Thread: Where's the mistake

1. ## Where's the mistake

I've made this exercise below. My broblem is that my solution doesn't match the solution in the book.Where did I make the mistake?

$sin ^{4} \frac{x}{3} +cos ^{4} \frac{x}{3}= \frac{5}{8}$

$\left(1-cos ^{2} \frac{x}{3} \right) ^{2}+cos ^{4} \frac{x}{3}= \frac{5}{8}$

$1-2cos ^{2} \frac{x}{3}+cos ^{4} \frac{x}{3}+cos ^{4} \frac{x}{3}= \frac{5}{8}$

$2cos ^{4} \frac{x}{3}-2cos ^{2} \frac{x}{3}=- \frac{3}{8}$

$cos ^{4} \frac{x}{3}-cos ^{2} \frac{x}{3}=- \frac{3}{16}$

$cos ^{4} \frac{x}{3}-cos ^{2} \frac{x}{3}+ \frac{3}{16}=0$

$\Delta=1- \frac{3}{4}= \frac{1}{4}$

$\sqrt{\Delta}= \frac{1}{2}$

$cos ^{2} \frac{x}{3}= \frac{3}{4} \vee cos ^{2} \frac{x}{3}= \frac{1}{4}$

$cos \frac{x}{3}= \frac{ \sqrt{3} }{2} \vee cos \frac{x}{3}=- \frac{ \sqrt{3} }{2} \vee cos \frac{x}{3}= \frac{1}{2} \vee cos \frac{x}{3}=- \frac{1}{2}$

$\frac{x}{3}= \frac{\Pi}{3}+2k\Pi \vee \frac{x}{3}=- \frac{\Pi}{3}+2k\Pi \vee \frac{x}{3}= \frac{\Pi}{6}+2k\Pi \vee \frac{x}{3}=- \frac{\Pi}{6}+2k\Pi$

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

This is my result.

Below it's the correct result:

$x=\Pi+3k\Pi \vee x=-\Pi+3k\Pi \vee x= \frac{\Pi}{2}+3k\Pi \vee x=- \frac{\Pi}{2}+3k\Pi$

2. Hello,

You're almost right.

The mistake is here :

$\frac{x}{3}= \frac{\Pi}{3}+2k\Pi \vee \frac{x}{3}=- \frac{\Pi}{3}+2k\Pi \vee \frac{x}{3}= \frac{\Pi}{6}+2k\Pi \vee \frac{x}{3}=- \frac{\Pi}{6}+2k\Pi
$
Remember that $\cos(x)=\cos(-x)$
so you should have been warned there's a problem here =)

$\cos \frac x3=-\frac{\sqrt{3}}{2}=\cos \frac{5 \pi}{6}$

Don't forget the $\cos(x)=\cos(-x)$ formula, which should give twice more solutions.

Now, you can shorten it.
In fact, you can notice that $\cos(x)=\cos(a) \implies x=a +2k \pi \vee x=-a+2k \pi$

$\cos^2(x)=\cos^2(a) \implies \cos(x)=\cos(a) \vee \cos(x)=-\cos(a)=\cos(a+\pi)$

Try to gather all of these, you'll get that :

$\cos^2(x)=\cos^2(a) \implies x=a+ k \pi \vee x=-a+k \pi$