# 3sin2

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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