`1/2 (2costhetacos3theta) cos2theta=1/4 rArr (cos2theta+ cos4theta) cos2theta=1/2`
`rArr 1.2[2cos^(2)2theta+2cos4thetacos2theta]=1/2 rArr 1+cos4theta+2cos4thetacos2theta=1`
`therefore cos4theta =0` or `(1+2cos2theta)=0`
Now from the first equation: `2cos4theta=0=cos(pi//2)`
`therefore 4theta=(n+1/2)pi rArr theta=(2n+1)pi/8, n in I`
for `n=0, theta=pi/8, n=1, theta=(3pi)/8, n=2, theta=(5pi)/8, n=3, theta=(7pi)/8` `(therefore 0 le thetalepi)` and from the second equation.
`cos2theta=-1/2=-cos(pi//3) = cos(pi-pi//3) = cos(2pi//3)`
`therefore 2theta=2kpi+-2pi//3 therefore theta=kpi+ pi//3, k in I`
Again for `k=0, theta=pi/3, k=1 , theta=(2pi)/(3) (therefore 0 le theta le pi)`
`therefore theta=pi/8, pi/3, (3pi)/8, (5pi)/8, (2pi)/8, (7pi)/8` Ans.
