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)ny, 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,


3sin2 x – 5 sin x cos x + 8 cos2 x = 2


3sin2 x + 3 cos2 x – 5sin x cos x + 5 cos2 x = 2


3 - 5sin x cos x + 5 cos2 x = 2 { sin2 x + cos 2x = 1 }


5cos2 x + 1 = 5sin x cos x


Squaring both sides:


(5cos2 x + 1)2 = (5sin x cos x)2


25cos4 x + 10cos2 x + 1 = 25 sin2 x cos2 x


25cos4 x + 10cos2 x + 1 = 25 (1 - cos2 x) cos2 x


50cos4 x – 15 cos2 x + 1 = 0


50cos4 x – 10 cos2 x – 5cos2 x + 1 = 0


10cos2 x ( 5cos2 x – 1) – (5cos2 x – 1) = 0


(10cos2 x - 1)(5cos2 x – 1) = 0


cos2 x = 1/10 or cos2 x = 1/5


Hence, when cos2 x = 1/10


We have,


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


let cos α = 1/√10


cos (π – α) = -1/√10


x = 2nπ ± α or x = 2nπ ± (π – α)


when,



When cos2 x = 1/5


We have, .


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


let cos β = 1/√5


cos (π – β) = -1/√5


x = 2mπ ± β or x = 2mπ ± (π – β)


when, .



ans


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