Solve the following equations :sec x cos 5x + 1 = 0, 0 < x < \(\frac{

1.	Solve the following equations :sec x cos 5x + 1 = 0, 0 < x < \(\frac{π}2\)
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, sec x cos 5x + 1 = 0, 0 < x < \(\frac{π}2\) ⇒ sec x cos 5x = -1 ⇒ cos 5x = - cos x ∵ - cos x = cos (π – x) ∴ cos 5x = cos (π – x) If cos x = cos y, implies 2nπ ± y, where n ∈ Z. ∴ 5x = 2nπ ± (π – x) ⇒ 5x = 2nπ + (π – x) or 5x = 2nπ – (π – x) ⇒ 6x = (2n+1)π or 4x = (2n-1)π ∴ x = (2n + 1) \(\frac{π}6\) or x = (2n - 1) \(\frac{π}4\)where n ∈ Z. But, 0 < x < \(\frac{π}2\) ∴ x = \(\frac{π}6\)and x = \(\frac{π}4\)....ans

Solve the following equations :sec x cos 5x + 1 = 0, 0 < x < \(\frac{π}2\)

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,

sec x cos 5x + 1 = 0, 0 < x < \(\frac{π}2\)

⇒ sec x cos 5x = -1

⇒ cos 5x = - cos x

∵ - cos x = cos (π – x)

∴ cos 5x = cos (π – x)

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

∴ 5x = 2nπ ± (π – x)

⇒ 5x = 2nπ + (π – x) or 5x = 2nπ – (π – x)

⇒ 6x = (2n+1)π or 4x = (2n-1)π

∴ x = (2n + 1) \(\frac{π}6\) or x = (2n - 1) \(\frac{π}4\)where n ∈ Z.

But, 0 < x < \(\frac{π}2\)

∴ x = \(\frac{π}6\)and x = \(\frac{π}4\)....ans