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