1.

Solve : `2cos^(2)x+3sinx=0`.

Answer» We have `2cos^(2)x+3sinx=0`
`rArr2(1-sin^(2)x)+3sinx=0`
`rArr2sin^(2)x-3sinx-2=0`
`rArr2sin^(2)x-4sinx+sinx-2=0`
`rArr2sinx(sinx-2)+(sinx-2)=0`
`rArr(sinx-2)(2sinx+1)=0`
`rArr(sinx-2)=0or(2sinx+1)=0`
`rArrsinx=2or(2sinx+1)=0`
`2sinx+1=0[becausesinx=2` is not possible]
`rArrsinx=-(1)/(2)=-"sin"(pi)/(6)=sin(pi+(pi)/(6))="sin"(7pi)/(6)`
`rArrsinx="sin"(7pi)/(6)`
`rArrx={npi+(-1)^(n)*(7pi)/(6)}`, where `ninI`.
Hence , the solution is given by `x={npi+(-1)^(n)*(7pi)/(6)}`, where `ninI`.


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