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

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

Explanation

Solution

First simplify double angle i.e. 2x2x in terms of single angle i.e. xx using the double angle formula for sine function which is given as follows:
2sin⁡xcos⁡x2\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⁡(2x)−sin⁡(x)=0\sin \left( {2x} \right) - \sin \left( x \right) = 0, we will first convert the double angle (2x)(2x) into single angle (x)(x) with the
help of double angle formula for sine function which is given as
2sin⁡xcos⁡x2\sin x\cos x
Therefore the given trigonometric equation will be written as
⇒sin⁡(2x)−sin⁡(x)=0 ⇒2sin⁡xcos⁡x−sin⁡x=0  \Rightarrow \sin\left( {2x} \right) - \sin \left( x \right) = 0 \\\ \Rightarrow 2\sin x\cos x - \sin x = 0 \\\
Now taking sin⁡x\sin x common in left hand side, we will get
⇒2sin⁡xcos⁡x−sin⁡x=0 ⇒sin⁡x(2cos⁡x−1)=0  \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(2cos⁡x−1)=0\sin x\left( {2\cos x - 1} \right) = 0 two possibilities are creating here for the solution of the given trigonometric equation,
⇒sin⁡x=0  or  (2cos⁡x−1)=0 ⇒sin⁡x=0  or  cos⁡x=12  \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  and  x=π3x = 0\;{\text{and}}\;x = \dfrac{\pi }{3} sine and cosine function have their respective values of 0  and  120\;{\text{and}}\;\dfrac{1}{2}
Therefore the required solution for the trigonometric equation sin⁡(2x)−sin⁡(x)=0\sin \left( {2x} \right) - \sin \left( x \right) = 0 is given by
x=0  and  x=π3x = 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\sin x = 0 is given as follows
x=±nπ,  where  n∈Ix = \pm n\pi ,\;{\text{where}}\;n \in I
And general solution of cos⁡x=12\cos x = \dfrac{1}{2} is given as
x=2nπ±π3,  where  n∈Ix = 2n\pi \pm \dfrac{\pi }{3},\;{\text{where}}\;n \in I
Therefore the general solution of the given trigonometric equation sin⁡(2x)−sin⁡(x)=0\sin \left( {2x} \right) - \sin \left( x \right) = 0 is given as
x=(±nπ∪2nπ±π3),  where  n∈Ix = \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.