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Thread: [SOLVED] Solve the equation

  1. #1
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    [SOLVED] Solve the equation

    Solve $\displaystyle \sin^4x+\sin^42x+\sin^43x=\cos^4x+\cos^42x+\cos^43 x$.
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  2. #2
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    Hello, james_bond!

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

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


    Factor:

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


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

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


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



    We have:

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

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

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