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Question

Solve cos3x+cos2x=sin(3x/2)+sin(x/2),0x<2π.

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Solution

cos3x+cos2x=sin(3x/2)+sin(x/2)
2cos5x2cosx22sinxcosx2=0
2cos(x/2)[cos(5x/2)sinx]=0
cos(x2)=0 x2=(n+12)π
or x=2nπ+π,
n=0, x=π as 0x2π
cos5x2=sinx=cos(π2x)
5x2=2nπ±(π2x)
Taking +ive sign (5x/2)+x=2nπ+π/2,
(7x/2)=(4n+1)(π/2)
x=(4n+1)π/7
n=0,x=π/7,n=1,x=5π/7,n=2,x=9π/7,
n=3,x=13π/7
Now taking ive sign,
(5x/2)x=2nππ/2
or (3x/2)=(4n1)π/2
x=(4n1)π/3,
For n=1, x=π.
x=π7,5π7,π,9π7,13π7.

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