Solve Trig Inequality: 2cos^2(x) + 1 = 3cos(2x) [0,2pi)

[lovemake1]

Homework Statement

solve the following equations or inequalties for x in the interval [0,2pi)

2cos^2(x) + 1 = 3cos(2x)

Homework Equations

The Attempt at a Solution

My attempt at the problem:

2cosx(cos2x) + 1 = 3cos(2x)
2cosx(cos^2x - sin^2x) + 1 = 3cos(2x)
2cos^3x - 2cosxsin^2x + 1 = 3(1-2sin^2x)
2cos^3x - 2cosxsin^2x + 1 = 3-6sin^2x
2cos^3x - 2cosxsin^2x + 1 - 3 + 6sin^2x = 0

and I am getting lost here...


please help,

 

[JonF]

use this identity:
Cos(2x) = 2Cos^2(x) - 1

 

[Mark44]

Homework Statement

solve the following equations or inequalties for x in the interval [0,2pi)

2cos^2(x) + 1 = 3cos(2x)


My attempt at the problem:

You have a mistake in your first step, below. 2cos²(x) is not equal to 2cos(x)*cos(2x). Use the identity that JonF gave to rewrite 3cos(2x) in terms of cos²(x).

2cosx(cos2x) + 1 = 3cos(2x)
2cosx(cos^2x - sin^2x) + 1 = 3cos(2x)
2cos^3x - 2cosxsin^2x + 1 = 3(1-2sin^2x)
2cos^3x - 2cosxsin^2x + 1 = 3-6sin^2x
2cos^3x - 2cosxsin^2x + 1 - 3 + 6sin^2x = 0

and I am getting lost here...


please help,