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If 5cos2theta+2"cos"^(2)(theta)/(2)+1=0,...

If `5cos2theta+2"cos"^(2)(theta)/(2)+1=0`, when `(0ltthetaltpi)`, then the values of `theta` are

A

`(pi)/(3)+-pi`

B

`(pi)/(3),cos^(-1)((3)/(5))`

C

`cos^(-1)((3)/(5))+-pi`

D

`(pi)/(3),pi-cos^(-1)((3)/(5))`

Text Solution

Verified by Experts

The correct Answer is:
D
Given equation is `5cos2theta+2"cos"^(2)(theta)/(2)+1=0`
`rArr5(2cos^(2)theta-1)+1+costheta+1=0`
`rArr10cos^(2)theta+costheta-3=0`
`rArr(2costheta-1)(5costheta+3)=0`
`rArrcostheta=(1)/(2)orcostheta=-(3)/(5)`
`rArrtheta=(pi)/(3)ortheta=pi-cos^(-1)((3)/(5))`
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