Question

sin(x)^4+ cos(x)^4=1/2

Answer 1

I'll assume that what was meant was $\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$.

The exponent in the funny place is just an abbreviation:   $\sin ^4 x = (\sin x)^4$.

I hope that's what you meant. Let me know if I'm wrong.

Let's start from the old saw

$\cos^2 x + \sin ^2x = 1$

Squaring both sides,

$(\cos^2 x + \sin ^2x)^2 = 1^2$

$\cos^4 x + 2 \cos ^2 x \sin ^2x +\sin ^4x = 1$

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

So now the original question 

$\sin ^4 x + \cos ^4 x = \dfrac{1}{2}$

becomes
$1 - 2 \cos ^2 x \sin ^2x = \dfrac{1}{2}$

$4 \cos ^2 x \sin ^2x = 1$

Now we use the sine double angle formula

$\sin 2x = 2 \sin x \cos x$

We square it to see

$\sin^2 2x = 4\sin^2 x \cos^2 x = 1$

Taking the square root,

$\sin 2x = \pm 1$

Not sure how you want it; we'll do it in degrees. 

When we know the sine of an angle, there's usually two angles on the unit circle that have that sine.  They're supplementary angles which add to $180^\circ$.  But when the sine is 1 or -1 like it is here, we're looking at $90^\circ$ and $-90^\circ$, which are essentially their own supplements, slightly less messy. 

That means we have two equations:

$\sin 2x = 1 = \sin 90^\circ$

$2x = 90^\circ + 360^\circ k \quad$ integer $k$

$x = 45^\circ + 180^\circ k$

or 


$\sin 2x = -1 = \sin -90^\circ$

$2x = -90^\circ+ 360^\circ k$

$x = - 45^\circ + 180^\circ k$

We can combine those for a final answer,

$x = \pm 45^\circ + 180^\circ k \quad$ integer $k$

Check.  Let's just check one, how about

$x=-45^\circ + 180^\circ = 135^\circ$

$\sin(135)= 1/\sqrt{2}$

$\sin ^4(135)=(1/\sqrt{2})^4 = 1/4$

$\cos ^4(135)=(-1/\sqrt{2})^4 = 1/4$

$\sin ^4(135^\circ) +\cos ^4(135^\circ) = 1/2 \quad\checkmark$