Q. 4 B5.0( 1 Vote )

# Solve the following equations :

cos x + cos 3x – cos 2x = 0

Answer :

**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,

cos x - cos 2x + cos 3x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, cos x - cos 2x + cos 3x = 0

∴ we will use cos x and cos 2x for transformation as after transformation it will give cos 2x term which can be taken common.

∴ cos x - cos 2x + cos 3x = 0

⇒ -cos 2x + (cos x + cos 3x) = 0

{∵ cos A + cos B = 2

⇒ -cos 2x + 2 cos

⇒ -cos 2x + 2cos 2x cos x = 0

⇒ cos 2x ( -1 + 2 cos x) = 0

∴ cos 2x = 0 or 1 + 2cos x = 0

⇒ cos 2x = cos π/2 or cos x = 1/2

⇒ cos 2x = cos π/2 or cos x = cos (π/3) = cos (π /3)

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

From above expression and on comparison with standard equation we have:

y = π/2 or y = π/3

∴ 2x = 2nπ ± π/2 or x = 2mπ ± π/3

**∴** **where m, n** **ϵ** **Z**

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