`sin3theta+cos2theta=0`
`rArr sin3theta=-cos2theta`
`rArr sin3theta=sin((3pi)/(2)-2theta)`
`rArr 3theta=npi+(-1)^(n).((3pi)/(2)-2theta)`
If n=2k, then
`3theta=2kpi+(-1)^(2k). ((3pi)/(2)-2theta)`
`rArr 3theta=2kpi+(-1)^(2k).((3pi)/(2)-2theta)`
`rArr 5theta=2kpi +(3pi)/(2)`
`rArr theta=1/5(2kpi+(3pi)/(2))` Ans.
If `n=2k+1`, then
`3theta=(2k+1)pi+(-1)^(2k+1)((3pi)/(2)-2theta)`
`rArr 3theta=2kpi+pi-((3pi)/(2)-2theta)` `[therefore (-1)^(2k+1)=-1]`
`rArr theta=(2kpi-pi/2)`. Ans.
