12sin^2+cos-6=0 I'm having trouble solving this the proper way. Please help. Thanks?

Sin^2 = 1 - cos^2

12[1-cos^2] + cos -6 =0

12 cos^2 -cos -6 =0

Solve this equation for cos as you would any quadratic equation.

Maybe the reason you are having trouble is because you don't even have a variable.

12sin^2 + cos - 6 = 0 ----> this makes no sense.

12sin^2x + cosx - 6 = 0 ----> now this makes sense.

Anyway, replace sin^2x with 1 - cos^2x to get an equation using only cosx

This will make it easier to solve, by factoring or using quadratic formula.

12 sin²x + cosx − 6 = 0

12 (1 − cos²x) + cosx − 6 = 0

−12 cos²x + cosx + 6 = 0

12 cos²x − cosx − 6 = 0

12 cos²x − 9 cosx + 8 cosx − 6 = 0

3 cosx (4 cosx − 3) + 2 (4 cosx − 3) = 0

(4 cosx − 3) (3 cosx + 2) = 0

cosx = 3/4 or −2/3

cosx = 3/4 ------> x = 2πk ± cos⁻¹(3/4)

cosx = −2/3 ----> x = 2πk ± cos⁻¹(−2/3)