Q. 4 B5.0( 1 Vote )

# Solve the following equations :cos x + cos 3x – cos 2x = 0

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ny, 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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