We're asked to solve,

\longrightarrow\sin x-3\sin(2x)+\sin(3x)=\cos x-3\cos(2x)+\cos(3x)⟶sinx−3sin(2x)+sin(3x)=cosx−3cos(2x)+cos(3x)

or,

\longrightarrow\sin(3x)+\sin x-3\sin(2x)=\cos(3x)+\cos x-3\cos(2x)⟶sin(3x)+sinx−3sin(2x)=cos(3x)+cosx−3cos(2x)

We have,

\sin A+\sin B=2\sin\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)sinA+sinB=2sin(

2

A+B

)cos(

2

A−B

)

\cos A+\cos B=2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)cosA+cosB=2COS(

2

A+B

)cos(

2

A−B

)

Then,

\footnotesize\text{$\longrightarrow2\sin\left(\dfrac{3x+x}{2}\right)\cos\left(\dfrac{3x-x}{2}\right)-3\sin(2x)=2\cos\left(\dfrac{3x+x}{2}\right)\cos\left(\dfrac{3x-x}{2}\right)-3\cos(2x)$}⟶2sin(

2

3x+x

)cos(

2

3x−x

)−3sin(2x)=2cos(

2

3x+x

)cos(

2

3x−x

)−3cos(2x)

\longrightarrow2\sin(2x)\cos x-3\sin(2x)=2\cos(2x)\cos x-3\cos(2x)⟶2sin(2x)cosx−3sin(2x)=2cos(2x)cosx−3cos(2x)

Taking \sin(2x)sin(2x) COMMON in LHS and \cos(2x)cos(2x) common in RHS,

\longrightarrow\sin(2x)(2\cos x-3)=\cos(2x)(2\cos x-3)⟶sin(2x)(2cosx−3)=cos(2x)(2cosx−3)

Taking RHS to LHS,

\longrightarrow\sin(2x)(2\cos x-3)-\cos(2x)(2\cos x-3)=0⟶sin(2x)(2cosx−3)−cos(2x)(2cosx−3)=0

\longrightarrow(\sin(2x)-\cos(2x))(2\cos x-3)=0⟶(sin(2x)−cos(2x))(2cosx−3)=0

This implies,

\longrightarrow2\cos x-3=0⟶2cosx−3=0

\longrightarrow \cos x=\dfrac{3}{2}⟶cosx=

2

3

This is not possible since \cos x\in\left[-1,\ 1\right]\not\cancel\ni\dfrac{3}{2}.cosx∈[−1, 1]



∋

2

3

.

And,

\longrightarrow\sin(2x)-\cos(2x)=0⟶sin(2x)−cos(2x)=0

\longrightarrow\sin(2x)=\cos(2x)⟶sin(2x)=cos(2x)

\longrightarrow\dfrac{\sin(2x)}{\cos(2x)}=1⟶

cos(2x)

sin(2x)

=1

\longrightarrow\tan(2x)=1⟶tan(2x)=1

\Longrightarrow 2x=n\PI+\dfrac{\pi}{4}⟹2x=nπ+

4

π

\longrightarrow\underline{\underline{x=\dfrac{n\pi}{2}+\dfrac{\pi}{8}}}⟶

x=

2

nπ

+

8

π

This is the SOLUTION of the equation where n\in\mathbb{Z}.n∈Z.