# How do you find the general solutions for cos (x)^4-sin (x)^4=2*cos (x)*cos(2x)?

Oct 19, 2015

There are 6 solutions over the interval $\left[0 , 2 \pi\right]$. The "general solution can be found by adding $2 \pi$ to each of those 6 solutions.

#### Explanation:

I'll exclude the variable x for simplicity:

${\cos}^{4} - {\sin}^{4} = 2 \cos \cos \left(2\right)$

Now simplify:

$\left({\cos}^{2} - {\sin}^{2}\right) \left({\cos}^{2} + {\sin}^{2}\right) = 2 \cos \left({\cos}^{2} - {\sin}^{2}\right)$

$\left({\cos}^{2} - {\sin}^{2}\right) \left[\left({\cos}^{2} + {\sin}^{2}\right) - 2 \cos\right] = 0$

$\left(\cos - \sin\right) \left(\cos + \sin\right) \left(1 - 2 \cos\right) = 0$

Now, we have 3 factors, when set equal to 0, will give us two unique solutions each for a total of 6 solutions over the interval $\left[0 , 2 \pi\right]$

$\cos - \sin = 0$; Solutions: $\frac{\pi}{4} , \frac{5 \pi}{4}$
$\cos + \sin = 0$; Solutions: $\frac{3 \pi}{4} , \frac{7 \pi}{4}$
$1 - 2 \cos = 0$; Solutions: $\frac{\pi}{3} , \frac{5 \pi}{3}$

Again, the general solution is simply adding/subtracting multiples of $2 \pi$ to each of these 6 solutions.

hope that helped