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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