Question

The general solution of \[\sin x-3\sin 2x+\sin 3x=\cos x-3cos2x+\cos 3x\] is
1) \[n\pi +\dfrac{\pi }{8}\]
2) \[\dfrac{n\pi }{2}+\dfrac{\pi }{8}\]
3) \[{{(-1)}^{n}}\dfrac{n\pi }{2}+\dfrac{\pi }{8}\] \[\]
4) \[2n\pi +{{\cos }^{-1}}\left( \dfrac{3}{2} \right)\]

Answer 1

Hint: In this type of question you need to first simplify the given equation in the question and try to reduce the given equation in the form of a trigonometric equation through which we can easily find the general solution that is in the form of a basic general trigonometric equation.

Complete step-by-step solution:
Here we will use the basic trigonometric identities to solve the question, by trigonometric identities what I mean is,
Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities which are true for every value occurring on both sides of an equation. Geometrically, these identities involve certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle. The trigonometric identities hold true only for the right-angle triangle.
As it is given in the question,
\[\sin x-3\sin 2x+\sin 3x=\cos x-3cos2x+\cos 3x\]
Now if we arrange above equation by simply changing the positions of specific terms we get,
\[\Rightarrow \sin x+\sin 3x-3\sin 2x=\cos x+\cos 3x-3\cos 2x\]
Now we will use the trigonometric identity and that is:
\[\sin c+\sin d=2\sin \left( \dfrac{c+d}{2} \right)\cos \left( \dfrac{c-d}{2} \right)\]
So after using above identity in above equation we get,
\[\Rightarrow 2\sin \left( \dfrac{x+3x}{2} \right)\cos \left( \dfrac{x-3x}{2} \right)-3\sin 2x=\cos x+\cos 3x-3\cos 2x\]
Now on further solving the expression we get,
\[\Rightarrow 2\sin \left( 2x \right)\cos \left( -x \right)-3\sin 2x=\cos x+\cos 3x-3\cos 2x\]
Now since we know that \[\cos (-x)=\cos (x)\]
So we have,
\[\Rightarrow 2\sin 2x\cos x-3\sin 2x=\cos x+\cos 3x-3\cos 2x\]
Now we will use the trigonometric identity and that is:
\[\cos c+\cos d=2\cos \left( \dfrac{c+d}{2} \right)\cos \left( \dfrac{c-d}{2} \right)\]
So after using above identity in above equation we get,
\[\Rightarrow 2\sin 2x\cos x-3\sin 2x=2\cos \left( \dfrac{x+3x}{2} \right)\cos \left( \dfrac{x-3x}{2} \right)-3\cos 2x\]
Similarly like above steps we get,
\[\Rightarrow 2\sin 2x\cos x-3\sin 2x=2\cos 2x\cos x-3\cos 2x\]
Now on taking common we get,
\[\Rightarrow \sin 2x(2\cos x-3)=\cos 2x(2\cos x-3)\]
Since we know that \[-1\le \cos x\le 1\]
Therefore,
 \[-5\le 2\cos x-3\le -1\]
So, \[2\cos x-3\ne 0\]
So, we can cancel the same terms on both sides,
Therefore now we left with,
\[\Rightarrow \sin 2x=\cos 2x\]
\[\Rightarrow \dfrac{\sin 2x}{\cos 2x}=1\]
 By using trigonometric identity we get,
\[\Rightarrow \tan 2x=1\]
Now it is a trigonometric equation. Now we need to solve this.
To solve the above we must know what trigonometric equations are and what general solutions of those equations tell us.
The expression involving integer \['n'\] this gives all solutions of a trigonometric equation. To derive general solution we will use the fact that:
Values of \[\sin x\] repeat after an interval of \[2\pi \] .
Values of \[\cos x\] repeat after an interval of \[2\pi \] .
Values of \[\tan x\] repeat after an interval of \[\pi \] .
So after using this concept of general solution in the equation we got:
\[\Rightarrow \tan 2x=1\]
We have,
\[\Rightarrow 2x=n\pi +\dfrac{\pi }{4}\]
\[\Rightarrow x=\dfrac{n\pi }{2}+\dfrac{\pi }{8}\]
So, the general solution of the given equation in question is
\[\Rightarrow x=\dfrac{n\pi }{2}+\dfrac{\pi }{8}\] .
Therefore the final answer is option$(2)$.

Note: Trigonometry can be used to roof a house, to make the roof inclined (in the case of single individual bungalows) and the height of the roof in buildings etc. It is used in the naval and aviation industries. It is used in cartography (creation of maps).It contributes in calculus also.