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Find the general solution of the equation sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x - Mathematics

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प्रश्न

Find the general solution of the equation sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x

बेरीज
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उत्तर

Given that: sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x

⇒ (sin3x + sinx) – 3sin2x = (cos3x + cosx) – 3cos2x

⇒ `2sin((3x + x)/2) . cos((3x - x)/2) - 3sin2x = 2cos((3x + x)/2).cos((3x - x)/2) - 3cos2x`

⇒ 2sin2x . cosx – 3sin2x = 2cos2x . cosx – 3cos2x

⇒ 2sin2x cosx – 2cos2x . cosx = 3sin2x – 3cos2x

⇒ 2cosx (sin2x – cos2x) = 3(sin2x – cos2x)

⇒ 2cosx(sin2x – cos2x) – 3(sin2x – cos2x) = 0

⇒ (sin2x – cos2x)(2cosx – 3) = 0

⇒ sin2x – cos2x = 0 and 2cosx – 3 ≠ 0   ....[∵ – 1 ≤ cos x ≤ 1]

⇒ `(sin2x)/(cos2x) - 1` = 0

⇒ tan2x = 1

⇒ tan2x = `tan  pi/4`

⇒ 2x = `npi + pi/4`

∴ x = `(npi)/2 + pi/8`

Hence, the general solution of the equation is x = `(npi)/2 + pi/8`, n ∈ Z.

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पाठ 3: Trigonometric Functions - Exercise [पृष्ठ ५५]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 3 Trigonometric Functions
Exercise | Q 28 | पृष्ठ ५५

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