Q19 of 76 Page 52

If sec x cos 5x + 1 = 0, where 0 < x ≤ π/2, then find the value of x.

Given,


sec x cos 5x = -1


⇒ cos 5x = -1/sec x


⇒ cos 5x + cos x = 0 {∵ sec x = 1/cos x}


By transformation formula of T-ratios we know that –


cos A + cos B =


⇒


⇒ 2 cos 3x cos 2x = 0


⇒ cos 3x = 0 or cos 2x = 0


∵ 0 < x ≤ π/2


∴ 0< 2x ≤ π or 0< 3x ≤ 3π/2


∴ 2x = π/2


⇒ x = π/4


3x = π/2


⇒ x = π/6


Or 3x = 3π/2


⇒ x = π/2


Hence, x = π/6, π/4.


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