# Math Help - Trigonometry

1. ## Trigonometry

Find all solutions on the interval 0 is less than or equal to x is less than 2π:
cos²x-cos(2x)=0

2. $\cos{A \pm B} = \cos{A}\cos{B} \mp \sin{A}\sin{B}$

$\cos{2x} = \cos{(x + x)}$

3. Hi nystudent2729,

We have $\cos^2(x)-\cos(2x)=0$ Note that $cos(2x) \equiv 2\cos^2(x)-1$ Thus we are left with

$\cos^2(x)-(2\cos^2(x)-1)=1-\cos^2(x)=0$

$\implies \cos^2(x)=1 \implies \cos(x)=\pm1$

Can you continue?