Answer :


Ideas required to solve the problem:


The general solution of any trigonometric equation is given as –


• sin x = sin y, implies x = nπ + (– 1)n y, where n Z.


• cos x = cos y, implies x = 2nπ ± y, where n Z.


• tan x = tan y, implies x = nπ + y, where n Z.


given,


2sin2 x +√3 cos x + 1 = 0


As the equation is of 2nd degree, so we need to solve a quadratic equation.


First we will substitute trigonometric ratio with some variable k and we will solve for k


As, sin2 x = 1 – cos2 x


we have,





Let, cos x = k


2k2 - √3 k – 3 = 0


2k2 -2√3 k + √3 k – 3 = 0


2k(k – √3) +√3(k – √3) = 0


(2k + √3)(k - √3) = 0


k = √3 or k = -√3/2


cos x = √3 or cos x = -√3/2


As cos x lies between -1 and 1


cos x can’t be √3


So we ignore that value.


cos x = -√3/2


cos x = cos 150° = cos 5π/6


If cos x = cos y, implies x = 2nπ ± y, where n Z.


On comparing our equation with standard form, we have


y = 5π/6


where n ϵ Z ..ans


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