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Let 0 < θ1 < θ2 < θ3 < denote the po...

Let `0 < θ_1 < θ_2 < θ_3 <` denote the positive solution of the equation `3+3costheta=2sin^2theta` dot The value of `theta_3+theta_7` is

A

`6pi`

B

`7pi`

C

`8pi`

D

`4pi`

Text Solution

Verified by Experts

The correct Answer is:
A

`3+3 cos theta =2 -2 cos^(2) theta`
`rArr 2 cos^(2) theta+3 cos theta+1=0`
`rArr (2 cos theta+1) (cos theta+1) =0`
`:. Cos theta=-1 or cos theta=-1/2`
If `cos theta=-1` then `theta=pi, 3 pi, 5 pi, 7 pi, 9pi`, ...
If `cos theta=-1//2` then `theta=(2pi)/3, (4pi)/3, (8pi)/3, (10 pi)/3, (14 pi)/3, ...`
Solution in increasing order are
`(2pi)/3 lt pi lt (4pi)/3 lt (8 pi)/3 lt 3pi lt (10 pi)/3 lt (14 pi)/3 lt 5pi`, ...
`:. theta_(3)+theta_(7)=(4pi)/3+(14 pi)/3=6pi`
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