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.