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