Trigonometry

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Can you please help me solve this trig equation for 0 ≤ x ≤ 2π?
9sin^2(x)-6cos(x)-10=0

Thank you!

  • Trigonometry -

    Turn it into a quadratic equation for u = cos x.

    9(1 - cos^2x)-6 cos x - 10 = 0

    -9u^2 -6u -1 = 0
    (3u +1)^2 = 0
    cos x = -1/3
    There will be solutions in the second and third quadrant

  • Trigonometry -

    Thanks.
    Am I supposed to use the inverse trig function to find the angle?

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