Question 478100: solve the equation sin^2 x=cos^2 (x/2) on the interval [0,2pi)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! solve the equation sin^2 x=cos^2 (x/2) on the interval [0,2pi)
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sin^2 x=cos^2 (x/2
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using half-angle formula for cosine
cos(x/2)=± sqrt((1+cos x)/2)
cos^2 (x/2)=(1+cos x)/2
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sin^2x=(1+cos x)/2
1-cos^2x=(1+cos x)/2
2-2 cos^2x=1+cos x
2 cos^2x+cos x-1=0
(2cos x-1)(cos x+1)=0
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2cos x-1=0
cos x=1/2
x=π/3, 5π/3
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cos x+1=0
cos x=-1
x=π
ans:
In the interval [0,2π]
x=π/3, π, and 5π/3
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