# How do I solve this equation; 0<x<2π(Pi) 9cos^2(x)-2cos(x)-3=0 answer must be in terms of π(Pi)I am not sure on how to even start anserwing this question, please explain the steps if...

How do I solve this equation;

0<x<2π(Pi) 9cos^2(x)-2cos(x)-3=0

answer must be in terms of π(Pi)

I am not sure on how to even start anserwing this question, please explain the steps if possible?

thank you

### 2 Answers | Add Yours

`9cos^2(x)-2cos(x)-3=0`

Let cosx = t

Then;

`9t^2-2t-3 = 0`

`t = [-(-2)+-sqrt((-2)^2-4*9*(-3))]/(2*9)`

So t = 0.699 OR t = -0.476

If t = 0.699;

cosx = 0.699

`cosx = cos 0.253pi`

General solution for cosines is `x = 2npi+-alpha` where `n in Z`

`x = 2npi+-0.253pi`

When n=-1 then x = -1.747pi OR x = -2.253pi

When n=-0 then x = 0.253pi OR x = -0.253pi

When n=-1 then x = 2.253pi OR x = 1.747pi

When n=-1 then x = 4.253pi OR x = 3.747pi

If t = -0.476

Cosx = -0.476

`Cosx = cos 0.342pi`

`x = 2mpi+-0.342pi` where `m in Z`

`x = 2mpi+-0.342pi`

When n=-1 then x = -1.658pi OR x = -2.342pi

When n=-0 then x = 0.342pi OR x = -0.342pi

When n= 1 then x = 2.747pi OR x = 1.658pi

But the answers should be x in (0,2pi)

So the answers are;

## `x = 0.253pi, 1.747pi, 0.342pi, 1.658pi`