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Mathematics Question on Trigonometric Equations

Solve the equation sinx−3sin2x+sin3x=cosx−3cos2x+cos3xsinx - 3sin2x + sin3x = cosx - 3cos2x + cos3x.

A

nπ2−π8,n∈I\frac{n\pi}{2}-\frac{\pi}{8}, n \in I

B

nπ2,n∈I\frac{n\pi}{2}, n \in I

C

nπ2+π8,n∈I\frac{n\pi}{2}+\frac{\pi}{8}, n \in I

D

nπ,n∈In\pi, n \in I

Answer

nπ2+π8,n∈I\frac{n\pi}{2}+\frac{\pi}{8}, n \in I

Explanation

Solution

Given sinx−3sin2x+sin3x=cosxsinx - 3sin2x + sin3x = cosx −3cos2x+cos3x- 3cos2x + cos3x ⇒(sin3x+sinx)−3sin2x=(cos3x+cosx)−3cos2x\Rightarrow \left(sin3x + sinx\right) - 3sin2x = \left(cos3x + cosx\right) - 3cos2x ⇒2sin2x cosx−3sin2x=2cos2xcosx−3cos2x\Rightarrow 2sin2x\, cosx - 3sin2x = 2cos2xcosx - 3cos2x ⇒sin2x(2cosx−3)=cos2x(2cosx−3)\Rightarrow sin2x\left(2cosx - 3\right) = cos2x\left(2cosx - 3\right) ⇒sin2x=cos2x(∵cos x≠32)\Rightarrow sin2x=cos2x\quad\left(\because cos\,x \ne\frac{3}{2}\right) ⇒tan2x=1\Rightarrow tan2x = 1 ⇒tan2x=tanπ4\Rightarrow tan2x=tan \frac{\pi}{4} ⇒2x=nπ+π4\Rightarrow 2x=n\pi+\frac{\pi}{4} ⇒x=nπ2+π8,n∈I\Rightarrow x=\frac{n\pi}{2}+\frac{\pi}{8}, n \in I.