Question 422033: I'm Stumped. Solve each equation on the indicated interval. Find EXACT answer
Sin(2Θ)Sin(Θ)= Cos(Θ), 0≤Θ<2π Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! Sin(2Θ)Sin(Θ)= Cos(Θ), 0≤Θ<2π
=Sin(2x)Sin(x)= Cos(x), 0≤x<2Pi
=2sinxcosxsinx=cosx
=2sin^2xcosx-cosx=0
=cosx(2sin^2x-1)=0
cosx=0
x=pi/2 or 3pi/2
..
2sin^2x-1=0
sin^2x=1/2
sin^2x-1/2=0
(sinx-1/sqrt2)(sinx+1/sqrt2)=0
sinx-1/sqrt2=0
sinx=1/sqrt2
x=pi/4 or 3pi/4
..
sinx+1/sqrt2=0
sinx=-1/sqrt2
x=5pi/4 or 7pi/4
..
ans:
x=pi/2 or 3pi/2
x=pi/4 or 3pi/4
x=5pi/4 or 7pi/4
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