Question

solve the equation for 0 less than and equal to x less than 2pi
sin^2x+cos2x-cosx=0

Answer 1

Re-arrange the equation as:\[\bf \sin^2(x)+\cos^2(x)=\cos(x)\]Now use the identity:
\[\bf \sin^2(x)+\cos^2(x)=1\]Substituting this we get:\[\bf \implies \cos(x)=1\]Now there is only two places in the interval \(\bf 0 \le x \le 2 \pi\) where \(\bf cos(x)=1\).Hence our solutions are:\[\bf x=0, 2 \pi\]
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