I'll assume that what was meant was [tex]\sin ^4 x + \cos ^4 x = \dfrac{1}{2}[/tex].
The exponent in the funny place is just an abbreviation: [tex]\sin ^4 x = (\sin x)^4[/tex].
I hope that's what you meant. Let me know if I'm wrong.
Let's start from the old saw
[tex]\cos^2 x + \sin ^2x = 1[/tex]
Squaring both sides,
[tex](\cos^2 x + \sin ^2x)^2 = 1^2[/tex]
[tex]\cos^4 x + 2 \cos ^2 x \sin ^2x +\sin ^4x = 1[/tex]
[tex]\cos^4 x + \sin ^4x = 1 - 2 \cos ^2 x \sin ^2x[/tex]
So now the original question
[tex]\sin ^4 x + \cos ^4 x = \dfrac{1}{2}[/tex]
becomes
[tex] 1 - 2 \cos ^2 x \sin ^2x = \dfrac{1}{2}[/tex]
[tex] 4 \cos ^2 x \sin ^2x = 1[/tex]
Now we use the sine double angle formula
[tex]\sin 2x = 2 \sin x \cos x[/tex]
We square it to see
[tex]\sin^2 2x = 4\sin^2 x \cos^2 x = 1[/tex]
Taking the square root,
[tex]\sin 2x = \pm 1[/tex]
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 [tex]180^\circ[/tex]. But when the sine is 1 or -1 like it is here, we're looking at [tex]90^\circ[/tex] and [tex]-90^\circ[/tex], which are essentially their own supplements, slightly less messy.
That means we have two equations:
[tex]\sin 2x = 1 = \sin 90^\circ[/tex]
[tex]2x = 90^\circ + 360^\circ k \quad[/tex] integer [tex]k[/tex]
[tex]x = 45^\circ + 180^\circ k[/tex]
or
[tex]\sin 2x = -1 = \sin -90^\circ[/tex]
[tex]2x = -90^\circ+ 360^\circ k[/tex]
[tex]x = - 45^\circ + 180^\circ k[/tex]
We can combine those for a final answer,
[tex]x = \pm 45^\circ + 180^\circ k \quad[/tex] integer [tex]k[/tex]
Check. Let's just check one, how about
[tex]x=-45^\circ + 180^\circ = 135^\circ[/tex]
[tex]\sin(135)= 1/\sqrt{2}[/tex]
[tex]\sin ^4(135)=(1/\sqrt{2})^4 = 1/4[/tex]
[tex]\cos ^4(135)=(-1/\sqrt{2})^4 = 1/4[/tex]
[tex]\sin ^4(135^\circ) +\cos ^4(135^\circ) = 1/2 \quad\checkmark[/tex]
