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If secx cos5x + 1 = 0, where 0 < x ≤ π2, then find the value of x.

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Question

If secx cos5x + 1 = 0, where 0 < x ≤ `pi/2`, then find the value of x.

Sum
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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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Chapter 3: Trigonometric Functions - Exercise [Page 54]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 3 Trigonometric Functions
Exercise | Q 19 | Page 54

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