Posted by **Lucy X** on Sunday, December 7, 2008 at 10:39pm.

Can you please help me solve this trig equation for 0 ≤ x ≤ 2π?

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

Thank you!

- Trigonometry -
**drwls**, Sunday, December 7, 2008 at 10:47pm
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 -
**Lucy X**, Sunday, December 7, 2008 at 10:52pm
Thanks.

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

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