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,


4sin2 x -8 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


4k2 + 8k – 5 = 0


4k2 -2k + 10k – 5 = 0


2k(2k – 1) +5(2k – 1) = 0


(2k + 5)(2k - 1) = 0


k = -5/2 = -2.5 or k = 1/2


cos x = -2.5 or cos x = 1/2


As cos x lies between -1 and 1


cos x can’t be -2.5


So we ignore that value.


cos x = 1/2


cos x = cos 60° = cos π/3


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


On comparing our equation with standard form, we have


y = π/3


where n ϵ Z ..ans


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