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
Can you please help me solve this trig equation for 0 ≤ x ≤ 2π?
9sin^2(x)-6cos(x)-10=0
Thank you!
2 answers
Thanks.
Am I supposed to use the inverse trig function to find the angle?
Am I supposed to use the inverse trig function to find the angle?
1 answer