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>>Solve for x: sin x + sin 2x...
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Solve for x:sinx+sin2x+sin3x=cosx+cos2x+cos3x in the interval 0≤x≤2π.
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Solution
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We write the given equation as
(sinx+sin3x)+sin2x=(cosx+cos3x)+cos2x
or 2sin2xcosx+sin2x=2cos2xcosx+cos2x
or sin2x(2cosx+1)=cos2x(2cosx+1)
or (sin2x−cos2x)(2cosx+1)=0
∴sin2x−cos2x=0 or 2cosx+1=0
If sin2x−cos2x=0, then tan2x=1,
Hence 2x=nπ+π/4
or x=(4n+1)
8
π
.(1)
If 2cosx+1=0, then cosx=−1/2 (2)
∴x=2nπ±
3
2π
or x=
3
6n±2
π
We seek values of x in the interval 0≤x≤2π
In this interval (1) gives
x=π/8,5π/8,9π/8,13π/8. (n=0,1,2,3)
and (2) gives x=2π/3,4π/3. (for n=0,1)
Thus we get the answer
x=π/8,5π/8,2π/3,9π/8,4π/3,13π/8.
