Solve the following equations : cos 4x = cos 2x

1.	Solve the following equations : cos 4x = cos 2x
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)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 4x = cos 2x If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. From above expression and on comparison with standard equation we have: y = 2x ∴ 4x = 2nπ ± 2x Hence, 4x = 2nπ + 2x or 4x = 2mπ - 2x ∴ 2x = 2nπ or 6x = 2mπ ⇒ x = nπ or x = \(\frac{2mπ}6\) = \(\frac{mπ}3\) ∴ x = nπ or \(\frac{mπ}3\)where m, n ϵ Z ..ans

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)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 4x = cos 2x

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

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

y = 2x

∴ 4x = 2nπ ± 2x

Hence,

4x = 2nπ + 2x or 4x = 2mπ - 2x

∴ 2x = 2nπ or 6x = 2mπ

⇒ x = nπ or x = \(\frac{2mπ}6\) = \(\frac{mπ}3\)

∴ x = nπ or \(\frac{mπ}3\)where m, n ϵ Z ..ans

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