If `sintheta,1,cos2theta` are in G.P., then find the general values of `theta`

Answer

`theta=npi+(-1)^(n-1) pi/2`

Explanation

`1^(2)= sin theta cos 2 theta`
or `1- sin theta (1-2 sin^(2) theta)=0`
or `2 sin^(3) theta- sin theta+1=0`
or `(sin theta+1) (2 sin^(2) theta-2 sin theta+1)=0`
or `sin theta=-1`
the other factor gives imaginary roots
`rArr theta=n pi +(-1)^(n) (-pi/2)`
`=n pi-(-1)^(n) pi/2= n pi+(-1)^(n-1) pi/2`