# Thread: [SOLVED] Solve the equation

1. ## [SOLVED] Solve the equation

Solve $\sin^4x+\sin^42x+\sin^43x=\cos^4x+\cos^42x+\cos^43 x$.

2. Hello, james_bond!

Solve: . $\sin^4\!x+\sin^4\!2x+\sin^4\!3x\;=\;\cos^4\!x+\cos ^4\!2x+\cos^4\!3x$

We have: . $(\cos^4\!x-\sin^4\!x) + (\cos^4\!2x-\sin^2\!2x) + (\cos^4\!2x-\sin^4\!3x) \;=\;0$

Factor:

$\underbrace{(\cos^2\!x-\sin^2\!x)}_{\text{This is }\cos2x}\underbrace{(\cos^2\!x+\sin^2\!x)}_{\text{ This is 1}} +$ $\underbrace{(\cos^2\!2x - \sin^2\!2x)}_{\text{This is }\cos4x}\underbrace{(\cos^2\!x+\sin^2\!x)}_{\text{ This is 1}} +$ $\underbrace{(\cos^2\!3x - \sin^2\!3x)}_{\text{This is }\cos6x}\underbrace{(\cos^2\!3x+\sin^2\!x)}_{\text {This is 1}} \;=\;0$

$\text{We have: }\;\cos4x + \underbrace{(\cos2x+\cos6x) }_{\text{sum-to-product}}\;=\;0$

. . . . . . . . $\cos4x + \overbrace{2\cos4x\cos2x} \;=\;0$

Factor: . $\cos4x(1 + 2\cos2x) \;=\;0$

We have:

. . $(a)\;\;\cos4x\:=\:0\quad\Rightarrow\quad 4x \:=\:\tfrac{\pi}{2} + \pi n \quad\Rightarrow\quad x \:=\;\tfrac{\pi}{8} + \tfrac{\pi}{4}n$

. . $(b)\;\;1+2\cos2x\:=\:0 \quad\Rightarrow\quad \cos2x\:=\:\text{-}\tfrac{1}{2} \quad\Rightarrow\quad 2x \:=\:\begin{Bmatrix}\frac{2\pi}{3}+2\pi n \\ \frac{4\pi}{3} + 2\pi n\end{Bmatrix} \quad\Rightarrow$ . $x \:=\:\begin{Bmatrix}\frac{\pi}{3} + \pi n \\ \frac{2\pi}{3} + \pi n\end{Bmatrix}$