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Math Help - solve for x: sinxcosx+0.5=sin^2 x

  1. #1
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    solve for x: sinxcosx+0.5=sin^2 x

    did I do this correctly so far?
    sinxcosx+0.5=sin^2 x
    -sin^2 x + sinxcosx = -0.5
    -sin x (sin x - cos x) = -0.5
    sin x (sin x - cos x) = 0.5

    If I did this correctly so far where do I go from here?
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  2. #2
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    Hello yoman360
    Quote Originally Posted by yoman360 View Post
    did I do this correctly so far?
    sinxcosx+0.5=sin^2 x
    -sin^2 x + sinxcosx = -0.5
    -sin x (sin x - cos x) = -0.5
    sin x (sin x - cos x) = 0.5

    If I did this correctly so far where do I go from here?
    This would be OK if you had zero on the RHS. The best way to go is to use the double-angle formulae: \sin x\cos x = \tfrac12\sin2x and \sin^2x = \tfrac12(1 - \cos2x)

    \sin x\cos x+0.5=\sin^2 x

    \Rightarrow \tfrac12\sin2x +0.5=\tfrac12(1 - \cos2x)

    \Rightarrow \sin2x +1=1 - \cos2x

    \Rightarrow \sin 2x +\cos2x =0

    Now use the fact that \sin(2x + \pi/4) = \sin2x\cos\pi/4 +\cos2x\sin\pi/4 = \tfrac{1}{\sqrt2}(\sin2x+\cos2x) and get:

    \sin(2x+\pi/4)=0

    \Rightarrow 2x +\pi/4 = 0, \pi, 2\pi, ...

    \Rightarrow x = -\pi/8, 3\pi/8, 7\pi/8, ...

    Grandad
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  3. #3
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    Hello, yoman360!

    Grandad is right . . . We need another approach.


    \sin x\cos x+\tfrac{1}{2}\:=\:\sin^2\!x

    We have: . \sin^2\!x - \sin x\cos x \:=\:\tfrac{1}{2}


    \text{Multiply by 2: }\;\underbrace{2\sin^2\!x}_{1-\cos2x} - \underbrace{2\sin x\cos x}_{\sin2x} \:=\:1 \quad\Rightarrow\quad 1 - \cos2x - \sin2x \:=\:1

    \text{We have: }\;\sin2x \:=\:-\cos2x \quad\Rightarrow\quad \frac{\sin 2x}{\cos2x} \:=\:-1

    Hence: . \tan2x \:=\:-1 \quad\Rightarrow\quad 2x \:=\:\frac{3\pi}{4}+\pi n \quad\Rightarrow\quad \boxed{x \:=\:\frac{3\pi}{8} + \frac{\pi}{2}n}

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