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The values of x1 between 0 and 2pi , sat...

The values of `x_1` between 0 and `2pi` , satisfying the equation `cos3x+cos2x=sin(3x)/2+sinx/2` are

A

6

B

7

C

4

D

5

Text Solution

Verified by Experts

The correct Answer is:
D
We have `cos3x+cos2x="sin"(3x)/(2)+"sin"(x)/(2)`
`rArr2"cos"(5x)/(2)"cos"(x)/(2)=2sinx"cos"(x)/(2)`
`[becausecosx+cosy=2" cos"(x+y)/(2)"cos"(x-y)/(2)]`
`sinx+siny=2"sin"(x+y)/(2)"cos"(x-y)/(2)`
Either `"cos"(x)/(2)=0rArr(x)/(2)=(2n+1)(pi)/(2)`
`rArrx=(2n+1)pior"cos"(5x)/(2)=sinx`
`rArr"cos"(5x)/(2)=cos((pi)/(2)-x)`
`rArr(5x)/(2)=2npi+-((pi)/(2)-x)`
Taking the (+ve)sign , `(7x)/(2)=2npi+(pi)/(2)rArrx=(4npi)/(7)+(pi)/(7)`
Taking the (-ve) sign,
`(3x)/(2)=2npi-(pi)/(2)rArrx=(4npi)/(3)-(pi)/(3) " For"0lexle2pi`
`x=(pi)/(7),(5pi)/(7),(9pi)/(7),(13pi)/(7),pi`
Thus ,the number of solutions is 5.
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