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
