Originally Posted by

**Sabo** I did get the right answers, but I still would like to make sure my thinking is accurate, and also check if I made unnecessary calculations somewhere.

Get all the solutions to the trigonometric equation $\displaystyle \sin^4 x + \cos^4 x = 1$ in the interval [0, 2pi[

What I did was:

$\displaystyle \sin x\sin x\sin x\sin x + \cos x\cos x\cos x\cos x = 1$

$\displaystyle

[-\dfrac{1}{2}(\cos 2x - \cos 0)][-\dfrac{1}{2}(\cos 2x - \cos 0)] + [\dfrac{1}{2}(\cos 2x + \cos 0)][\dfrac{1}{2}(\cos 2x + \cos 0)] = 1$

$\displaystyle

[-\dfrac{1}{2}(2\cos^2 x - 2)][-\dfrac{1}{2}(2\cos^2 x - 2)] + [\dfrac{1}{2}(2\cos^2 x)][\dfrac{1}{2}(2\cos^2 x)] = 1$

$\displaystyle (-\cos^2 x - 1)^2 + \cos^4 x = 1$

$\displaystyle 2\cos^4 x - 2\cos^2 x + 1 = 1$

$\displaystyle \cos^4 x - \cos^2 = 0$

We set u = cos^2 and get $\displaystyle u^2 - u = 0$ which gives us u = 0 and u = 1.

cos^2 x = 0 gives us cos x = +/- 0. x = pi/2 + k(pi).

cos^2 x = 1 gives us cos x = +/- 1. x = 0 + k(pi).

So the solutions are, in order,

0, pi/2, pi, 3pi/2.

I apologize for the mix and match between Latex and regular text.