If `5cos2 theta+2 "cos"^(2) (theta)/2+1=0`, when `(0 lt thetaltpi)` the the value of `theta` are

Options

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)`

Answer

D

Explanation

Given equation is
`5cos 2theta+2"cos"^(2)(theta)/2+1=-0`
`implies5(2cos^(2)theta-1)+1+costheta+1=0`
`implies10cos^(2)theta+cos-3=0`
`implies(2cos theta-1)(5 cos theta+3)=0`
`impliescos theta=1/2`
or `cos theta=-3/5impliestheta=(pi)/3`
or `theta=pi-cos^(-1)(3/5)`