Solve the following equations : sin^2 x - cos x = 1/4

Question

Solve the following equations :

sin² x - cos x = \(\frac{1}4\)

Answer 1

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,

sin²x - cos x = \(\frac{1}4\)

As the equation is of 2^nd 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,

1 - cos²x - cos x = \(\frac{1}4\)

⇒ 4 - 4 cos²x - 4 cos x = 1

⇒ 4 cos²x + 4 cos x - 3 = 0

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 = 1/2 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 = 1/2

⇒ 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 = \(\fracπ{3}\)

∴ x = 2nπ ± \(\fracπ{3}\)where n ϵ Z ..ans