Question

How do you find the exact solutions of the equation $\sin 2x - \sin x = 0$ in the interval $[0,2\pi )$ ?

Answer 1

Hint: In order to solve the equation, we substitute $\sin 2x$ with $2\sin x\cos x$ and thus we have the equation as: $2\sin x\cos x - \sin x = 0$
From here, we take out the common factor, and equate each factor with zero and solve further to get our required answer.

Complete step-by-step solution:
In the given, we have the equation: $\sin 2x - \sin x = 0$. In order to solve it, we need to replace the identity $\sin 2x$ with its expanded form.
We know that: $\sin 2x = 2\sin x\cos x$, therefore placing this value in the given equation, we get-
$ \Rightarrow 2\sin x\cos x - \sin x = 0$
Taking out the common factor from this equation, we get:
$ \Rightarrow \sin x\left( {2\cos x - 1} \right) = 0$
Now we have two separate factors: $\sin x$ and $2\cos x - 1$.
Let’s equate both these factors separately with $0$
Equating the first factor $\sin x$ with $0$, we get:
$ \Rightarrow \sin x = 0$
As we know that the value of $\sin x = 0$, when $x = {0^ \circ },\pi ,2\pi $ (according to the picture given below)

Now, let’s equate the other factor $2\cos x - 1$ with $0$
Thus, we have: $2\cos x - 1 = 0$
On adding $ + 1$ to both the sides, we get:
$2\cos x = 1$
On dividing both sides with $2$ , we get:
$\cos x = \dfrac{1}{2}$
Now, we know that $\cos x = \dfrac{1}{2}$ when $x = \pm \dfrac{\pi }{3}$
On referring to the picture above we find that $ - \dfrac{\pi }{3} = \left( {2\pi - \dfrac{{5\pi }}{3}} \right) = \dfrac{{5\pi }}{3}$

Thus the values are: $0,\dfrac{\pi }{3},\pi ,\dfrac{{5\pi }}{3},2\pi $

Note: Trigonometric identities are simply the equations which are true for right angled triangles. We may even consider a complete circle and divide it into four quadrants to help us understand our trigonometric identities as so:
 When the whole turn around the circle is equal to $2\pi $, while a half circle is equal to $\pi $ . The different quadrants are divided into different angles.
The first quadrant has angles from ${0^ \circ } - \dfrac{\pi }{2}$ , the second quadrant has angles form $\dfrac{\pi }{2}$ to $\pi $ , the third quadrant has angles from $\pi $ to $\dfrac{{3\pi }}{2}$ and the fourth quadrant has angles from $\dfrac{{3\pi }}{2}$ to $2\pi $