––– Solve :-
sin x − 3 sin2x + sin3x = cos x − 3 cos2x + cos3x
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Answers
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.