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Solve: `sin^(2)theta+2costheta+1/4=0`

Answer» `sin^(2)theta+2costheta+1/4=0`
`4sin^(2)theta+8costheta+1=0`
`rArr 4(1-cos^(2)theta)+8costheta+1=0`
`rArr 4cos^(2)theta-8costheta-5=0`
`rArr 4cos^(2)theta-10costheta+2costheta-5=0`
`rArr 2costheta(2costheta-5)+1(2costheta-5)=0`
`rArr (2costheta-5)(2costheta+1)=0`
`rArr 2costheta-5=0` or `2costheta+1=0`
Now `2costheta-5=0`
`rArr costheta=-1/2=-cospi/3`
`=cos(pi-pi/3)=cos(2pi)/(3)`
`therefore theta=2npi +- (2pi)/(3)` Ans.


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