Question

Solve the following equations : 

4 sin² x – 8 cos x + 1 = 0

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, 

4sin² x - 8 cos x + 1 = 0 

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,

Let, cos x = k 

∴ 4k² + 8k – 5 = 0 

⇒ 4k² - 2k + 10k – 5 = 0 

⇒ 2k(2k – 1) +5(2k – 1) = 0 

⇒ (2k + 5)(2k - 1) = 0 

∴ k = \(\frac{-5}2\) = -2.5 or k = \(\frac{1}2\)

⇒ cos x = -2.5 or cos x = \(\frac{1}2\)

As cos x lies between -1 and 1 

∴ cos x can’t be -2.5 

So we ignore that value. 

∴ cos x = \(\frac{1}2\)

⇒ cos x = cos 60° = cos \(\frac{π}3\)

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

On comparing our equation with standard form, we have 

y = \(\frac{π}3\)

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