Question

#QualityQuestion
@Trigonometry

What is the general solution of sin x − 3 sin2x + sin3x = cos x − 3 cos2x + cos3x.​

Answer 1

#quality answer

@trigonometry

Step-by-step explanation:

We have, (sin x + sin 3x) – 3 sin 2x = (cos x + cos 3x) – 3 cos 2x

=> 2 sin 2x cos x – 3 sin 2x = 2 cos 2x.cos x – 3 cos 2x

=> sin 2x(2 cos x – 3) = cos 2x(2 cos x – 3)

=> sin 2x = cos 2x (As cos x ≠ 3/2)

=> tan 2x = 1

=> tan 2x = tan π/4

=> 2x = nπ + π/4 ,n∈Z

x = nπ/2 +π/8 , n∈

Answer 2

~Solution :-

• The solution for your question is as follows;

$\sf{sinx - 3 sin2x + sin3x = cosx - 3 cos2x + cos3x}$

$\to \sf{ 2 sin2x \: cosx - 3 sin2x - 2 cos2x cosx + 3 cos2x = 0}$

$\to \sf{sin2x (2cosx - 3) - cos2x (2 cosx - 3) = 0 }$

$\to \sf{ (sin2x - cos2x) (2 cosx - 3) = 0}$

$\to \sf{ sin2x = cos2x}$

$\to \sf{ 2x = 2n \pi \pm( \frac{ \pi }{2 - 2x})}$

$\implies \bf \red{x = \frac{ nπ}{ 2} + \frac{π}{ 8}}$

• Hence, the general solution for the trigonometric identity will be $\sf{{x = \frac{ nπ}{ 2} + \frac{π}{ 8}}}$.

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