Question

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

Answer 1

`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`