**Question:**

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

**Solution:**

According to the question,

sec x cos 5x = -1

⇒ cos 5x = -1/sec x

We know that,

sec x = 1/cos x

⇒ cos 5x + cos x = 0

By transformation formula of T-ratios,

We know that,

$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$

$\Rightarrow 2 \cos \left(\frac{5 x+x}{2}\right) \cos \left(\frac{5 x-x}{2}\right)=0$

⇒ 2 cos 3x cos 2x = 0

⇒ cos 3x = 0 or cos 2x = 0

∵ 0 < x ≤ π/2

Therefore, 0< 2x ≤ π or 0< 3x ≤ 3π/2

Therefore, 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, π/2.