Question

How do you solve \[\sin 2x + \sin x = 0\] over the interval \[0\] to \[2\pi \]?

Answer 1

Hint: In this question, we have the trigonometric function. Which must solve for the interval\[0\]to\[2\pi \]. In the left-hand side, a trigonometric function \[\sin 2x\] will be there. To solve the above trigonometric equation, we use the formula. And the formula is given as below.
\[ \Rightarrow \sin 2a = 2\sin a\cos a\]

Complete step by step answer:
In this question, an equation of \[x\]is given, which I want to solve. First we know about an equation, an equation is defined as it has two things which are equal. And the equation also likes a statement “this equal that”. The equation has two things or two sides, the left side is known as the left hand side and the right side is known as right hand side. The left hand side is denoted as “LHS” and the right hand side is denoted as “RHS”.
Now, come to the question. The equation is given below.
\[ \Rightarrow \sin 2x + \sin x = 0\]
First, we take the left hand side from the above equation and want to solve it.
Then, the left hand side is.
\[\sin 2x + \sin x\]
We know that, \[\sin 2a = 2\sin a.\cos a\]
Then above the left hand side is written as below.
\[ \Rightarrow \sin 2x + \sin x = 2\sin x\cos x + \sin x\]
Now we write right hand side is equal to right hand side.
Then,
\[2\sin x.\cos x + \sin x = 0\]
We solve the above equation as below.
We take the \[\sin x\] as common from the left hand side.
Then,
\[\sin x\left( {2\cos x + 1} \right) = 0\]
We know that, if the product of any number of terms is equal to zero then one of the terms must equal to zero. Then
\[
  \sin x = 0 \\
  2\cos x + 1 = 0 \\
 \]
Now, we find the value of \[x\]for the interval\[0\]to\[2\pi \].
Then,
\[\sin x = 0\]
               For the interval\[0\]to\[2\pi \], the value of\[x\]is.
              \[x = 0,\;x = \pi \]And\[x = 2\pi \]
\[2\cos x + 1 = 0\]
              \[
  2\cos x = - 1 \\
  \cos x = - \dfrac{1}{2} \\
 \]
For the interval\[0\]to\[2\pi \], the value of\[x\]is.
\[x = \dfrac{{2\pi }}{3}\] And \[x = \dfrac{{4\pi }}{3}\]

Therefore, the value of \[x\] for the interval \[0\] to \[2\pi \] be \[0,\;\pi ,\;\dfrac{{2\pi }}{3},\;\dfrac{{4\pi }}{3},\;2\pi \].

Note:
If you have an equation that you want to solve. Then first you separate the left hand side and right hand side. Then take them one (left hand side or right hand side) and solve it. After taking the other side and solving it. Then write the left hand side is equal to the right hand side. And then solve that for its variable.