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Math Help - Solving a trigonometric equation --> help needed!

  1. #1
    s3a
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    Solving a trigonometric equation --> help needed!

    Can someone help me with fully illustrated steps for the following please?:

    Solve the following trigonometric equation. Give the exact value(s).

    2 sin^2 (x) - 3 cos (x) = 3 x ∈ [0, π]


    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by s3a View Post
    Can someone help me with fully illustrated steps for the following please?:

    Solve the following trigonometric equation. Give the exact value(s).

    2 sin^2 (x) - 3 cos (x) = 3 x ∈ [0, π]


    Any help would be greatly appreciated!
    Thanks in advance!
    2(1-cos^2(x))-3cos(x)=3

    2-2cos^2(x)-3cos(x)-3=0

    -2cos^2(x)-3cos(x)-1=0

    (-2cosx-1)(cosx+1)=0

    so the first is zero or the second find when the first zero and when the second zero
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  3. #3
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    Hello, s3a!

    Solve the following trigonometric equation. Give the exact value(s).

    . . 2\sin^2\!x - 3\cos x \:=\:3 \qquad x \in [0, \pi]

    \text{We have: }\;2\underbrace{\sin^2\!x} - 3\cos x \;=\;3
    . . . . . 2\overbrace{(1-\cos^2\!x)} - 3\cos x \;=\;3

    . . which simplifies to: . 2\cos^2\!x + 3\cos x + 1 \:=\:0

    . . which factors: . (\cos x + 1)(2\cos x + 1) \:=\:0


    And we have two equations to solve:

    . . \cos x +1\:=\:0 \quad\Rightarrow\quad \cos x \:=\:-1 \quad\Rightarrow\quad\boxed{ x \:=\:\pi}

    . . 2\cos x + 1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:-\tfrac{1}{2} \quad\Rightarrow\quad\boxed{ x \:=\:\tfrac{2\pi}{3}}

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