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Solve the equation `(sqrt(3))/2sinx-cosx=cos^2x`

Answer» `sqrt3/2sinx=cos^2x+cosx`
`sqrt3sinx=2(cos^2x+cosx)`
`3sin^2x=4(cos^2x+cosx)^2`
`3(1-cos^2x)=4(cos^4x+2cos^3x+cos^2x)`
`4cos^4x+8cos^3x+7cos^2x-3=0`
`(cosx+1)(2cosx-1)(2cos^2x+3cosx+3)=0`
`cosx=-1,1/2`
`x=(2n+1)pi,2npi+pi/3`.


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