Cos2x=cos^2x-1/2
i think you mean solve \[\cos(2x)=\cos^2(x)-\frac{1}{2}\]
do you want to prove the equality?
first note that \[\cos(2x)=\cos^2(x)-\sin^2(x)\] That means that the equation becomes \[\cos^2(x)-\sin^2(x) = \cos^2(x)-\frac{1}{2}\] \[\sin^2(x) = \frac{1}{2}\] \[\sin(x) = \frac{\sqrt{2}}{2}\] the principle value is \[x=\frac{\pi}{4}\] however note that sinx must be positive so only angles in quadrants 1 and 2 will work. So the answer is \[x=\frac{\pi}{4}+2n\pi\] or \[x=\frac{3\pi}{4}+2n\pi\] where n is a integer. because sin(3pi/4) also equals sqrt(2)/2 also note that adding an integer multiple of twopi will not change the answer.
@allism1995 its an invalid expression
no he wants to solve the equation for x
@stylestar5678
ya ik
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