solve:
cos^2(x) + cos(x) = cos2x

2. Originally Posted by chopstixdcb
solve:
cos^2(x) + cos(x) = cos2x
Note that $\cos(2x)=2\cos^2x-1$.

Therefore, your equation becomes $\cos^2x-\cos x-1=0$.

Letting $u=\cos x$, we end up with the quadratic equation $u^2-u-1=0$.

Can you take it from here?

3. so now, all i have to do is factor it right?

4. Originally Posted by chopstixdcb
so now, all i have to do is factor it right?
If possible; Otherwise, resort to quadratic formula.

Also, remember this is a quadratic in terms of $\cos x$, so once you have your solutions, you need to solve $\cos x=u_1$ and $\cos x=u_2$ to find the solutions that you're seeking.

5. oh geez...i feel so dumb right now. okay thanks a lot.

6. Originally Posted by Chris L T521
If possible; Otherwise, resort to quadratic formula.

Also, remember this is a quadratic in terms of $\cos x$, so once you have your solutions, you need to solve $\cos x=u_1$ and $\cos x=u_2$ to find the solutions that you're seeking.
umm, if you don't mind, can you show me the final steps. for some reason i got non real solution and the other one is cosx=1

7. $u^2 - u - 1 = 0$

$u^2 - u + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - 1 = 0$

$\left(u - \frac{1}{2}\right)^2 - \frac{5}{4} = 0$

$\left(u - \frac{1}{2}\right)^2 = \frac{5}{4}$

$u - \frac{1}{2} = \frac{\pm \sqrt{5}}{2}$

$u = \frac{1 \pm \sqrt{5}}{2}$.

Note that $-1 \leq \cos{x} \leq 1$ for all $x$. Are there any solutions that don't fit?