# Trigonometry

• Aug 31st 2008, 08:52 AM
nystudent2729
Trigonometry
Find all solutions on the interval 0 is less than or equal to x is less than 2π:
cos²x-cos(2x)=0
• Aug 31st 2008, 09:11 AM
Chop Suey
$\cos{A \pm B} = \cos{A}\cos{B} \mp \sin{A}\sin{B}$

$\cos{2x} = \cos{(x + x)}$
• Aug 31st 2008, 09:12 AM
Sean12345
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?