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 = (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 = (1)/(2)`

becomes

`1 - 2 \cos ^2 x \sin ^2x = (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/√(2)`


`\sin ^4(135)=(1/√(2))^4 = 1/4`


`\cos ^4(135)=(-1/√(2))^4 = 1/4`


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