How do you solve 3cos^2 x - 5cos x -1= 3 sin^2 x?

1 Answer

x = 2npi +- 2pi/3

The description looks long, but I've written every single step so that anyone can understand!

Explanation:

3cos^2x−5cosx−1=3sin^2x

3(cos^2x - sin^2x) - 5cosx-1=0

3(cos^2x-1+cos^2x) - 5cosx -1=0

6cos^2x -5cosx -4=0

from here its just a quadratic equation. you divide it to its factors.

(3cosx-4) (2cosx +1) =0

3cosx-4=0 or 2cosx+1=0
from here u get two possibilities

cos x = 4/3 or cos x = -1/2
the first answer cannot be true because the cosines should be between -1 and +1
if u remember the cosine curve u should be able to understand what im talking about.

so we are left with the latter answer

cosx= -pi/3
cos x = (pi - pi/3) you dont really have to show this step.. u can do it in your mind ;)
cos x =2pi/3

according to the trig equations theta = 2npi +- theta
just replace the correct places of this equations with our answer
so the answer is
x = 2npi +- 2pi/3