Q. 3 A5.0( 2 Votes )

Solve the following equations :


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,



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 + 4k – 3 = 0


4k2 -2k + 6k – 3


2k(2k – 1) +3(2k – 1) = 0


(2k – 1)(2k + 3) = 0


k =1/2 or k = -3/2


cos x = � or cos x = -3/2


As cos x lies between -1 and 1


cos x can’t be -3/2


So we ignore that value.


cos x = �


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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