Problem on Trigonometric Equations

Topic : Trigonometric Equation

Problem : Solve 3Sin θ Cos θ – 2 Cos θ = 0 and 0 ≤ θ ≤ 2π

Solution :

3Sin θ Cos θ – 2 Cos θ = 0
Taking Cos θ as common
Cos θ (3Sin θ - 2) = 0
Cos θ = 0 or 3Sin θ – 2 = 0
3 Sin θ – 2 = 0
3Sin θ = 2
Sin θ = 2/3 (2/3 = 0.66 = 41.8º ~ 42º)
Sin θ = Sin 42º
θ = n π + (-1)ⁿ 42º
when n = 0 , θ = 42º
when n = 1 , θ = π - 42º = 138º

Let’s solve Cos θ = 0
Cos θ = Cos π/2
The formula for Cosine is :
If Cos x = Cos y
Then x = 2nπ ± y

So here it will be θ = 2nπ ± π/2
When n = 0, θ = 2(0)π ± π/2 = ± π/2
But only + π/2 lies in 0 ≤ θ ≤ 2π
So x = π/2

When n = 1, θ = 2(1)π ± π/2 = 2π ± π/2 = 2π ± π/2 = (4π ± π)/2 = 3π/2, 5π/2
But x = 3π/2

Now the further values of n will give angles greater than 360º

So θ will be 42º, 90º, 138º, 270º