Solve `2 cos^(2) theta +3 sin theta=0`.

Explanation

We have,
`2cos^(2)theta+3 sintheta=0`
`2cos^(2)theta+3sintheta=0`
`2(1-sin^(2)theta)+3sintheta=0`
`2sin^(2)theta-4sintheta+sintheta-2=0`
`2sintheta(sintheta-2) + 1(sintheta-2)=0`
`rArr 2sintheta + 1=0` [Since `sintheta ne 2 therefore sintheta-2 ne 0]`
`therefore sintheta=-1/2`
`rArr sintheta =sin(-pi/6)`
`rArr theta=npi+(-1)^(n) (-pi/6), n int Z`
`rArr theta=npi + (-1)^(n) (-pi/6), n int Z`
`rArr theta=npi +(-1)^(n+1)pi/6, n int Z`