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How do you solve $\sin \left( {2x} \right) - \sin \left( x \right) = 0?$

seo-qna
Last updated date: 13th Jun 2024
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Answer
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Hint: First simplify double angle i.e. $2x$ in terms of single angle i.e. $x$ using the double angle formula for sine function which is given as follows:
$ 2\sin x\cos x$. Then further simplify and find the principal solution and then the general solution.

Complete step by step solution:
To solve the given trigonometric expression $\sin \left( {2x} \right) - \sin \left( x \right) = 0$, we will first convert the double angle $(2x)$ into single angle $(x)$ with the
help of double angle formula for sine function which is given as
$ 2\sin x\cos x$
Therefore the given trigonometric equation will be written as
$
\Rightarrow \sin\left( {2x} \right) - \sin \left( x \right) = 0 \\
\Rightarrow 2\sin x\cos x - \sin x = 0 \\
$
Now taking $\sin x$ common in left hand side, we will get
$
\Rightarrow 2\sin x\cos x - \sin x = 0 \\
\Rightarrow \sin x\left( {2\cos x - 1} \right) = 0 \\
$
Now from the final equation, that is $\sin x\left( {2\cos x - 1} \right) = 0$ two possibilities are creating here for the solution of the given trigonometric equation,
$
\Rightarrow \sin x = 0\;{\text{or}}\;\left( {2\cos x - 1} \right) = 0 \\
\Rightarrow \sin x = 0\;{\text{or}}\;\cos x = \dfrac{1}{2} \\
$
We know that at $x = 0\;{\text{and}}\;x = \dfrac{\pi }{3}$ sine and cosine function have their respective values of $0\;{\text{and}}\;\dfrac{1}{2}$
Therefore the required solution for the trigonometric equation $\sin \left( {2x} \right) - \sin \left( x \right) = 0$ is given by
$x = 0\;{\text{and}}\;x = \dfrac{\pi }{3}$
Hence we will now write its general solution,
We know that the general solution of $\sin x = 0$ is given as follows
$x = \pm n\pi ,\;{\text{where}}\;n \in I$
And general solution of $\cos x = \dfrac{1}{2}$ is given as
$x = 2n\pi \pm \dfrac{\pi }{3},\;{\text{where}}\;n \in I$
Therefore the general solution of the given trigonometric equation $\sin \left( {2x} \right) - \sin \left( x \right) = 0$ is given as
$x = \left( { \pm n\pi \cup 2n\pi \pm \dfrac{\pi }{3}} \right),\;{\text{where}}\;n \in I$


Note: The double angle formula is a special case of the addition formula of sine angle in which we consider both the arguments equal. Normally periodic functions have three types of solution that are principal solution which is the smallest possible solution, particular solution which lies according to the conditions given in the problem and general solution which is set of each and every possible solution.