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,
2sin2 x +√3 cos x + 1 = 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
As, sin2 x = 1 – cos2 x
∴ we have,
⇒
⇒
Let, cos x = k
∴ 2k2 - √3 k – 3 = 0
⇒ 2k2 -2√3 k + √3 k – 3 = 0
⇒ 2k(k – √3) +√3(k – √3) = 0
⇒ (2k + √3)(k - √3) = 0
∴ k = √3 or k = -√3/2
⇒ cos x = √3 or cos x = -√3/2
As cos x lies between -1 and 1
∴ cos x can’t be √3
So we ignore that value.
∴ cos x = -√3/2
⇒ cos x = cos 150° = cos 5π/6
If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.
On comparing our equation with standard form, we have
y = 5π/6
∴ where n ϵ Z ..ans