Question

How do you solve $\sin 2x - \sin x = 0$ ?

Answer 1

Hint:
Use the identity of the trigonometry, $\sin 2x = 2\sin x\cos x$ . Take $\sin x$ common from the whole equation. Then making the equation one by one equal to zero and finding the values for $x$ in radians or degrees.

Complete step by step solution:
We have given with the equation, $\sin 2x - \sin x = 0$. In this equation, we have to find the values for $x$.
Now, we know that, $\sin 2x = 2\sin x\cos x$
Using the above identity of $\sin 2x$ in the equation given in the question, we get –
$ \Rightarrow 2\sin x\cos x - \sin x = 0$
In the above equation, we have to simplify the equation to solve for $x$. Therefore, taking $\sin x$ common from the above equation, we get –
$ \Rightarrow \sin x\left( {2\cos x - 1} \right) = 0$
Now, making $\sin x$ and $2\cos x - 1$ one by one equal to zero, we get –
$
   \Rightarrow \sin x = 0 \\
   \Rightarrow x = {\sin ^{ - 1}}0 \\
 $
We know that the value of $\sin $ is equal to 0 when the angle is $0$. Therefore, one of the values of $x$ is equal to zero.
Now, making \[\left( {2\cos x - 1} \right)\] equal to zero, we get –
$ \Rightarrow 2\cos x - 1 = 0$
Shifting $1$ to another side in the above equation, we get –
$ \Rightarrow 2\cos x = 1$
Now, by using the transposition method we have to shift 2 on another side of the equation then, the function of operation of 2 changes to multiplication from division, we get –
$
   \Rightarrow \cos x = \dfrac{1}{2} \\
   \Rightarrow x = {\cos ^{ - 1}}\dfrac{1}{2} \\
 $
For finding the inverse of cosine for the value of $\dfrac{1}{2}$ , the angle we get is ${60^ \circ }$.
Hence, the other value of $x$ is ${60^ \circ }$.
The above value of $x$ is in degree. To convert the degree into radians, we multiply the angle in degrees with \[\dfrac{\pi }{{{{180}^ \circ }}}\]
Hence, the angle is $ \Rightarrow {60^ \circ } \times \dfrac{\pi }{{{{180}^ \circ }}}$
Therefore, the angle in radians is $\dfrac{\pi }{3}$.

Hence, the two values of $x$ are $0$ and ${60^ \circ }$ or $\dfrac{\pi }{3}$.

Note:
The trigonometric formula $\sin 2x = 2\sin x\cos x$ that we used in the question must be kept in mind. In this question, the students can check their answer by putting the values of $x$ in the equation given in the question and if the solution comes as 0 then, the values of $x$ are correct.