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Math Help - Relatively confused...very confused. Solving trig equations, work shown.

  1. #1
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    Relatively confused...very confused. Solving trig equations, work shown.

    1) cos(2θ)+6sinēθ=4 and
    2)cosθ=sinθ

    1) cos(2θ)+6sinēθ=4
    cos(2θ)-4=-6(1-cosēθ)
    cos(2θ)-4=-6+cosēθ
    -cosēθ+cos2θ=-2
    -cosēθ+cos2θ+2=0
    -cosēθ+2cosēθ+1=0
    cosθ(-cosθ+2cosθ)+1=0
    -cosθ+2cosθ+1=0
    cosθ+1=0
    cosθ=-1
    and, as you can imagine, that's not the answer in the back of the book :P
    2) cosθ=sinθ
    cosθ=(1-cosēθ)
    cosθ+cosēθ=1
    and I'm stuck there.

    Any help would be appreciated - I'm having considerable trouble with these
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Voluntarius Disco View Post
    1) cos(2θ)+6sinēθ=4 and
    2)cosθ=sinθ

    1) cos(2θ)+6sinēθ=4
    cos(2θ)-4=-6(1-cosēθ)
    cos(2θ)-4=-6+6cosēθ
    -cosēθ+cos2θ=-2
    -cosēθ+cos2θ+2=0
    -cosēθ+2cosēθ+1=0
    cosθ(-cosθ+2cosθ)+1=0

    -cosθ+2cosθ+1=0
    cosθ+1=0
    cosθ=-1
    and, as you can imagine, that's not the answer in the back of the book :P
    2) cosθ=sinθ
    cosθ=(1-cosēθ)
    cosθ+cosēθ=1
    and I'm stuck there.

    Any help would be appreciated - I'm having considerable trouble with these
    2) Use the fact that \frac{sin \theta}{cos \theta} = tan \theta

    Also note that sin(x) \neq 1-cos^2(x) but rather sin(x) = \sqrt{1-cos^2(x)}

    1) From where you were last correct

    cos(2\theta)-4 = -6+6cos^2 \theta

    2cos^2\theta-5 = -6+6cos^2\theta

    4cos^2 \theta -1 = 0

    Use the difference of two squares

    (2cos \theta -1)(2cos \theta +1) = 0
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  3. #3
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    Hello, Voluntarius Disco!

    1)\;\cos2\theta + 6\sin^2\theta \:=\:4

    . . \underbrace{\cos2\theta} + 6\sin^2\theta \:=\:4

    \overbrace{1-2\sin^2\theta} + 6\sin^2\theta \;=\;4 \quad\Rightarrow\quad 4\sin^2\!\theta \:=\:3 \quad\Rightarrow\quad \sin^2\!\theta \:=\:\frac{3}{4}

    . . \sin\theta \;=\;\pm\frac{\sqrt{3}}{2} \quad\Rightarrow\quad \theta \;=\;\begin{Bmatrix}\dfrac{\pi}{3} + \pi n \\ \\[-3mm] \dfrac{\pi}{3} + \pi n \end{Bmatrix}




    2)\;\cos\theta \:=\:\sin\theta

    We have: . \sin\theta \;=\;\cos\theta

    Divide by \cos\theta\!:\;\;\frac{\sin\theta}{\cos\theta} \;=\;\frac{\cos\theta}{\cos\theta} \quad\Rightarrow\quad \tan\theta \;=\;1

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

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