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Math Help - Trigonometric equation

  1. #1
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    Trigonometric equation

    Solve the equation:
    (\sin x+\cos x)\cdot \sin {2x}=\sqrt{2}.
    Or if you like the equivavalent:
    \sin x\cos x\cdot(\sin x+\cos x)=\frac{\sqrt{2}}2.
    Last edited by james_bond; December 30th 2008 at 03:45 AM.
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  2. #2
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    Dear james_bond,

    or if you like the equivavalent:
    [(1/\sqrt{2})sin(x)+(1/\sqrt{2})cos(x)]sin(2x)=1

    hint: in [] brackets there is sin(a+b) with suitable a and b.
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  3. #3
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    Hello, james_bond!

    Solve the equation: . (\sin x+\cos x)\cdot \sin {2x}\:=\:\sqrt{2}

    Square both sides: . (\sin x + \cos x)^2\sin^2\!2x \:=\:2

    . . . . . . \left(\sin^2\!x + 2\sin x\cos x + \cos^2\!x\right)\sin^2\!2x \:=\:2

    . . . . . . (\underbrace{2\sin x\cos x}_{\text{This is }\sin2x} + \underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}})\sin^2\!2x \;=\;2

    . . . . . . . . . . . . . . . (\sin2x + 1)\sin^2\!2x \:=\:2

    . . . . . . . . . . . . . . \sin^3\!2x + \sin^2\!2x - 2 \:=\:0

    \text{Factor: }\;\underbrace{(\sin2x-1)}_{\downarrow}\underbrace{\left(\sin^2\!2x + 2\sin2x + 2\right)}_{\text{No real roots}} \:=\:0

    \text{Then: }\:\sin2x \:=\:1 \quad\Rightarrow\quad2x \:=\:\tfrac{\pi}{2} + 2\pi n \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{4}+ \pi n


    But we find that half of the roots are extraneous.
    The equation is satisfied only if x is in Quadrant 1.

    Therefore: . x \;=\;\frac{\pi}{4} + 2\pi n

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