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Thread: little help please

  1. #1
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    little help please

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

    2[(sinx)^4 + (cosx)^4) =1
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  2. #2
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    Quote Originally Posted by white_cap View Post
    within the domain : (-p # x #p ), solve for x

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

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

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

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

    -Dan
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  3. #3
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    Hello, white_cap!

    Another approach . . .


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


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

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

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

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

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

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