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sin2x+cosx=0...

`sin2x+cosx=0`

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Given equation:
`sin2x+cosx=0`,
or `2sinxcosx+cosx=0`,
or `cosx(2sinx+1)=0`
`rArr cosx=0` or `2sinx+1=0`
If `cosx=0` then `x=(2n+1)pi/2`, where `n in z`, and `2sinx+1=0, rArr sinx=-1/2=sin(pi+pi/6)`
`rArr sinx=sin(pi + pi/6)`
`rArr sinx=sin(7pi)/(6)`
`rArr x=npi+(-1)^(n)(7pi)/(6)`, where `n in Z`
Therefore, the general solution of given equation is
`x=npi+(-1)^(n)(7pi)/(6)` or `(2n+1)pi/2, n in Z` Ans.
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