within the domain : (-p # x #p ), solve for x

2[(sinx)^4 + (cosx)^4) =1

2. Originally Posted by white_cap
within the domain : (-p # x #p ), solve for x

2[(sinx)^4 + (cosx)^4) =1
Your domain got screwed up a bit. Anyway:
$2(sin^4(x) + cos^4(x)) = 1$

$2(sin^4(x) + cos^2(x) \cdot cos^2(x) ) = 1$

$2(sin^4(x) + (1 - sin^2(x))(1 - sin^2(x))) = 1$
etc.

Write this out as a polynomial in sin(x) and solve.

-Dan

3. Hello, white_cap!

Another approach . . .

Solve for $x\!:\;\;\sin^4\!x + \cos^4\!x \:=\:1\quad -\pi \leq x \leq \pi$
Divide by 2: . $\sin^4\!x + \cos^4\!x \:=\:\frac{1}{2}$

Add and subtract $2\sin^2\!x\cos^2\!x\!:$
. . $\sin^4\!x \,{\color{blue}+\, 2\sin^2\!x\cos^2\!x} + \cos^4\!x \,{\color{blue}-\, 2\sin^2\!x\cos^2\!x} \;=\;\frac{1}{2}$

. . $(\sin^2\!x\; +\; \cos^2\!x)^2\;-\;\frac{1}{2}(4\sin^2\!x\cos^2\!x) \;=\;\frac{1}{2}\qquad\Rightarrow\qquad 1^2\;-\;\frac{1}{2}(2\sin x\cos x)^2 \;=\;\frac{1}{2}$

And we have: . $1 - \frac{1}{2}\sin^2\!2x \;=\;\frac{1}{2}\qquad\Rightarrow\qquad \sin^2\!2x \;=\;1\qquad\Rightarrow\qquad\sin2x \;=\;\pm1$

. . Hence: . $2x \;=\;-\frac{3\pi}{2},\;-\frac{\pi}{2},\;\frac{\pi}{2},\;\frac{3\pi}{2}$

Therefore: . $\boxed{x \;=\;-\frac{3\pi}{4},\;-\frac{\pi}{4},\;\frac{\pi}{4},\;\frac{3\pi}{4}}$