1.

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

Answer» `2(1-sin^2x) + 3sinx = 0`
`-2sin^2x + 3sinx + 2=0`
`2sin^2x - 3sinx - 2=0`
let `sinx = t`
`2t^2 - 3t -2 = 0`
`(2t +1)(t-2) = 0`
`t = -1/2 or t=2`
`sin x = -1/2 or sinx = 2`
solution is not possible for `sinx = 2`
`:. sinx = - sin(pi/6) = sin(-pi/6)`
`x = npi + (-1)^n(-pi/6) ; n in z`
answer


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