Asked by Lucy X

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

Thank you!
Answered by drwls
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
Answered by Lucy X
Thanks.
Am I supposed to use the inverse trig function to find the angle?
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