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Math Help - Precalculus Help. Finding exact solutions on the equation

  1. #1
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    Precalculus Help. Finding exact solutions on the equation

    Hi

    Find all exact solutions on [0,2pi)

    2(cos^2)x + 3sinx = 3

    Someone said i had to use (cos^2)x + (sin^2)x = 1 and use sin x as a variable, but i do not understand what that means.

    Can someone explain to me step by step?
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  2. #2
    Senior Member eumyang's Avatar
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    Exactly what that person said. Use sin x as a variable.

    Replace \cos^2 x with 1 - \sin^2 x:
    \begin{aligned}<br />
2\cos^2 x + 3\sin x &= 3 \\<br />
2(1 - \sin^2 x) + 3\sin x &= 3 \\<br />
2 - 2\sin^2 x + 3\sin x &= 3 \\<br />
2\sin^2 x - 3\sin x + 1 &= 0<br />
\end{aligned}

    See how this looks like a quadratic? If you replaced \sin x with y you would get
    2y^2 - 3y + 1 = 0

    Solve the quadratic for y, plug back in \sin x for y, and solve for x.
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  3. #3
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    thank you!

    so i factored out 2(sin^2)x - 3 sinx + 1 = 0

    i got (sinx -1) (sin x - 1)

    set them equal to zero and got sin x = 1

    making x = pi/2
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  4. #4
    Senior Member eumyang's Avatar
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    Quote Originally Posted by danielh9103 View Post
    so i factored out 2(sin^2)x - 3 sinx + 1 = 0

    i got (sinx -1) (sin x - 1)
    No, the factoring is not right (unless there is a typo)? If you multiply it out you get
    (\sin x - 1)(\sin x - 1) = \sin^2 x - 2\sin x + 1
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  5. #5
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    im sorry! it was a typo!

    (2 sinx -1) (sinx - 1)

    sinx= 1/2 sinx = 1

    x = pi/2, pi/6, 5pi/6
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  6. #6
    Senior Member eumyang's Avatar
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    Much better now!
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