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13 April, 04:13

Solve the equation 2cos^2x + 3sinx = 3

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  1. 13 April, 05:06
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    x = 2 π n_1 + π/2 for n_1 element Z

    or x = 2 π n_2 + (5 π) / 6 for n_2 element Z or x = 2 π n_3 + π/6 for n_3 element Z

    Step-by-step explanation:

    Solve for x:

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

    Write 2 cos^2 (x) + 3 sin (x) = 3 in terms of sin (x) using the identity cos^2 (x) = 1 - sin^2 (x):

    -1 + 3 sin (x) - 2 sin^2 (x) = 0

    The left hand side factors into a product with three terms:

    - (sin (x) - 1) (2 sin (x) - 1) = 0

    Multiply both sides by - 1:

    (sin (x) - 1) (2 sin (x) - 1) = 0

    Split into two equations:

    sin (x) - 1 = 0 or 2 sin (x) - 1 = 0

    Add 1 to both sides:

    sin (x) = 1 or 2 sin (x) - 1 = 0

    Take the inverse sine of both sides:

    x = 2 π n_1 + π/2 for n_1 element Z

    or 2 sin (x) - 1 = 0

    Add 1 to both sides:

    x = 2 π n_1 + π/2 for n_1 element Z

    or 2 sin (x) = 1

    Divide both sides by 2:

    x = 2 π n_1 + π/2 for n_1 element Z

    or sin (x) = 1/2

    Take the inverse sine of both sides:

    Answer: x = 2 π n_1 + π/2 for n_1 element Z

    or x = 2 π n_2 + (5 π) / 6 for n_2 element Z or x = 2 π n_3 + π/6 for n_3 element Z
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