Question

Solve the following equation: sin mx + sin nx = 0

Answer 1

Hint: In order to find the solution of this question, we should have some knowledge about a few trigonometric addition formulas like \[\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right).\] So, we will use this property to get the general solution of this question.

Complete step-by-step answer:
In this question, we have been asked to find the solution of sin mx + sin nx = 0. To solve this question, we should know about a few trigonometric identities like \[\sin x+\sin y=2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right).\] So for, x = mx and y = nx, we get,
\[\sin mx+\sin nx=2\sin \left( \dfrac{mx+nx}{2} \right)\cos \left( \dfrac{mx-nx}{2} \right)\]
So, by using the above equation, we can write the given equation as,
\[\Rightarrow 2\sin \left( \dfrac{mx+nx}{2} \right)\cos \left( \dfrac{mx-nx}{2} \right)=0\]
Now, we know that to satisfy the above equation, we can say,
\[\Rightarrow \sin \left( \dfrac{mx+nx}{2} \right)=0;\cos \left( \dfrac{mx-nx}{2} \right)=0\]
Now, we know that the general solution of \[\sin \theta =0\] is given by \[\theta =n\pi \] and the general solution of \[\cos \theta =0\] is given by \[\theta =n\pi +\dfrac{\pi }{2}.\] So, we can say the above two equalities can be written as,
\[\dfrac{mx+nx}{2}=k\pi ;\dfrac{mx-nx}{2}=k\pi +\dfrac{\pi }{2}\]
And we can further write it as,
\[\Rightarrow \dfrac{\left( m+n \right)x}{2}=k\pi ;\dfrac{\left( m-n \right)x}{2}=k\pi +\dfrac{\pi }{2}\]
\[\Rightarrow x=\dfrac{2k\pi }{m+n};x=\left( k\pi +\dfrac{\pi }{2} \right)\dfrac{2}{m-n}\]
\[\Rightarrow x=\dfrac{2k\pi }{m+n};x=\dfrac{2k\pi }{m-n}+\dfrac{\pi }{m-n}\]
Hence, we can say that the solution of sin mx + sin nx = 0 are \[x=\dfrac{2k\pi }{m+n}\text{ or }x=\dfrac{2k\pi }{m-n}+\dfrac{\pi }{m-n}.\]

Note: While solving the question, the possible mistake one can make is by finding the principal value instead of finding the general value because we haven’t been asked to find the principal value. So, we have to find the generalized value to get our answer. Also, we need to remember that the general solution of \[\sin \theta =0\text{ is }\theta =n\pi \text{ and }\cos \theta =0\text{ is }\theta =n\pi +\dfrac{\pi }{2}.\]