Solve the following equations :
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,
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
As, sin2 x = 1 – cos2 x
∴ we have,
⇒
⇒
Let, cos x = k
∴ 4k2 + 4k – 3 = 0
⇒ 4k2 -2k + 6k – 3
⇒ 2k(2k – 1) +3(2k – 1) = 0
⇒ (2k – 1)(2k + 3) = 0
∴ k =1/2 or k = -3/2
⇒ cos x = � or cos x = -3/2
As cos x lies between -1 and 1
∴ cos x can’t be -3/2
So we ignore that value.
∴ 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