Solving linear and quadratic Trig Equations

[anonymous12]

Homework Statement

Solve each equation for 0≤\Theta≤2\pi (exact values where possible)
The question:
2cos^22\Theta - cos2\Theta - 1=0

Homework Equations

cos2\Theta = 2cos^2\Theta-1

The Attempt at a Solution

2cos^22\Theta - 2cos^2\Theta - 1 - 1 = 0
2cos^22\Theta - 2cos^2\Theta - 2 = 0
2(cos^22\Theta - cos^2\Theta - 1)=0
I have absolutely no idea what to do after this.

4. Answer to the question
0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi

 

[danago]

Think about the equation as a quadratic equation in cos(2\theta). If it helps, make the substitution u=cos(2\theta) and then solve for u (you should get two values), from which you can then find theta.

 

[epenguin]

This is a possible attack that could give you the answers in a 'by inspection' sort of way, but I think you have misapplied this

cos2Θ=2cos2Θ−1

and this

2cos²2Θ−2cos2Θ−1−1=0

is wrong.

If you do it right I guess you will see the given answers are right.

However would that have worked for you if you didn't know the answers? More secure might be to solve the quadratic equation. (Or it might be slightly easier to get it all in cosθ and cos²θ and solve that, I don't know. Doing both would be a check.)