Solve: `sin3theta=sin2theta`

Explanation

`sin3theta=sin2theta`
`rArr 3theta=npi+(-1)^(n).2theta`
If n=2k, then
`3theta=2kpi+(-1)^(2k).2theta`
`=2kpi+2theta` `[therefore (-1)^(2k)=1]`
`rArr theta=2kpi` Ans.
If `n=2K+1`, then
`3theta=(2k+1)pi+(-1)^(2k+1).2theta`
`=(2k+1)pi-2theta` `[therefore (-1)^(2k+1)=-1]`
`rArr 5theta=(2k+1)pi`
`rArr theta =(2k+1)pi/5`. Ans.