Find all solutions on the interval 0 is less than or equal to x is less than 2π:

cos²x-cos(2x)=0

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- Aug 31st 2008, 08:52 AMnystudent2729Trigonometry
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 AMChop Suey
$\displaystyle \cos{A \pm B} = \cos{A}\cos{B} \mp \sin{A}\sin{B}$

$\displaystyle \cos{2x} = \cos{(x + x)}$ - Aug 31st 2008, 09:12 AMSean12345
Hi nystudent2729,

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

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

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

Can you continue?