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Question
If secx cos5x + 1 = 0, where 0 < x ≤ `pi/2`, then find the value of x.
Solution
secx cos5x = –1
⇒ cos5x = `(-1)/secx`
We know that
secx = `1/cosx`
⇒ cos5x + cosx = 0
By transformation formula of T-ratios,
We know that
cosA + cosB = `2cos(("A" + "B")/2) cos(("A" - "B")/2)`
⇒ `2cos((5x + x)/2) cos((5x - x)/2)` = 0
⇒ 2cos3x cos2x = 0
⇒ cos3x = 0 or cos2x = 0
∵ 0 < x ≤ `pi/2`
Therefore, 0 < 2x ≤ π or 0 < 3x ≤ `(3pi)/2`
Therefore, 2x = `pi/2`
⇒ x = `pi/4`
3x = `pi/2`
⇒ x = `pi/6`
Or 3x = `(3pi)/2`
⇒ x = `pi/2`
Hence, x = `pi/6, pi/4, pi/2`.
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