Solve the following equations :
2 cos2 x – 5 cos x + 2 =0
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,
2 cos2 x – 5 cos x + 2 =0
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
Let, cos x = k
∴ 2k2 – 5k + 2 = 0
⇒ 2k2 – 4k – k +2 = 0
⇒ 2k(k – 2) -1(k -2) = 0
⇒ (k – 2)(2k - 1) = 0
∴ k = 2 or k = �
⇒ cos x = 2 {which is not possible} or cos x = � (acceptable)
∴ 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