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Math Help - trigo prob

  1. #1
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    trigo prob

    find the most general value of angle A SUCH THAT

    3-2cosA-4sinA-cos2A+sin2A=0
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  2. #2
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    Hello, prasum!

    \text{Find the most general value of angle }x\text{ such that:}

    . 3-2\cos x - 4\sin x-\cos2x+\sin2x\:=\:0

    Using Double-angle Identities, we have:

    . . . 3 - 2\cos x - 4\sin x - (1 - 2\sin^2\!x) + 2\sin x\cos x \:=\:0

    . . . . . . . 2\sin^2\!x - 4\sin x + 2 - 2\cos x + 2\sin x\cos x \;=\;0

    Divide by 2:. . \sin^2\!x - 2\sin x + 1 + \sin x\cos x - \cos x \;=\;0

    Factor: . . . . . . . . . . . (\sin x - 1)^2 + \cos x(\sin x - 1) \;=\;0

    Factor:. . . . . . . . . . . . (\sin x - 1)(\sin x - 1 + \cos x) \;=\;0


    Then we have:

    [1]\;\sin x - 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:1 \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{2} + 2\pi n


    [2]\;\sin x -1 + \cos x \:=\:0 \quad\Rightarrow\quad \sin x + \cos x \:=\:1

    Divide by \sqrt{2}\!:\;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{\sqrt{2}}

    We have: . \cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x \;=\;\frac{1}{\sqrt{2}}

    . . . . . . . . . . . . . . . . \sin\left(x + \frac{\pi}{4}\right) \;=\;\frac{1}{\sqrt{2}}

    . . . . . . . . . . . . . . . . . . . . x + \frac{\pi}{4} \;=\;\begin{Bmatrix}\frac{\pi}{4} + 2\pi n \\ \frac{3\pi}{4} + 2\pi n \end{Bmatrix}

    . . . . . . . . . . . . . . . . . . . . . . . x \;=\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}

    \text{Solution: }\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}\;\text{ for any integer }n

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, prasum!


    Using Double-angle Identities, we have:

    . . . 3 - 2\cos x - 4\sin x - (1 - 2\sin^2\!x) + 2\sin x\cos x \:=\:0

    . . . . . . . 2\sin^2\!x - 4\sin x + 2 - 2\cos x + 2\sin x\cos x \;=\;0

    Divide by 2:. . \sin^2\!x - 2\sin x + 1 + \sin x\cos x - \cos x \;=\;0

    Factor: . . . . . . . . . . . (\sin x - 1)^2 + \cos x(\sin x - 1) \;=\;0

    Factor:. . . . . . . . . . . . (\sin x - 1)(\sin x - 1 + \cos x) \;=\;0


    Then we have:

    [1]\;\sin x - 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:1 \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{2} + 2\pi n


    [2]\;\sin x -1 + \cos x \:=\:0 \quad\Rightarrow\quad \sin x + \cos x \:=\:1

    Divide by \sqrt{2}\!:\;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{\sqrt{2}}

    We have: . \cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x \;=\;\frac{1}{\sqrt{2}}

    . . . . . . . . . . . . . . . . \sin\left(x + \frac{\pi}{4}\right) \;=\;\frac{1}{\sqrt{2}}

    . . . . . . . . . . . . . . . . . . . . x + \frac{\pi}{4} \;=\;\begin{Bmatrix}\frac{\pi}{4} + 2\pi n \\ \frac{3\pi}{4} + 2\pi n \end{Bmatrix}

    . . . . . . . . . . . . . . . . . . . . . . . x \;=\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}

    \text{Solution: }\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}\;\text{ for any integer }n

    thanks
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