# Solving linear and quadratic Trig Equations

1. Nov 15, 2011

### anonymous12

1. The problem statement, all variables and given/known data
Solve each equation for $$0≤\Theta≤2\pi$$ (exact values where possible)
The question:
$$2cos^22\Theta - cos2\Theta - 1=0$$

2. Relevant equations

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

3. 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.

$$0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$$

Last edited: Nov 15, 2011
2. Nov 15, 2011

### 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.

3. Nov 15, 2011

### 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.)