Question

How do you solve $ {\sin ^2}x + \sin x = 0 $ and solve the equation on the interval of $ \left( {0,2\pi } \right) $ ?

Answer 1

Hint: In order to determine the solution of the above trigonometric equation, pull out $ \sin x $ as common from the both terms to make the equation in factored form. Now equate every factor equal to zero to derive the required solution in the interval $ \left( {0,2\pi } \right) $ .

Complete step by step solution:
We are given a trigonometric equation $ {\sin ^2}x + \sin x = 0 $ and we have to find its solution
 $ {\sin ^2}x + \sin x = 0 $
Pulling out $ \sin x $ as common from the terms, we get
  $ \sin x\left( {\sin x + 1} \right) = 0 $ -------(1)
To find the solution for the above , we have to divide the both sides of the equation with $ \sin x $ and $ \left( {\sin x + 1} \right) $ one by one
First Dividing both sides of equation with $ \left( {\sin x + 1} \right) $ , we have
 \[
  \dfrac{1}{{\left( {\sin x + 1} \right)}} \times \sin x\left( {\sin x + 1} \right) = 0 \times \dfrac{1}{{\left( {\sin x + 1} \right)}} \\
  \sin x = 0 \\
  x = {\sin ^{ - 1}}\left( 0 \right) \;
 \]
The value of $ x $ is an angle having sine value 0. Since we know $ \sin 0 = 0 \to 0 = {\sin ^{ - 1}}\left( 0 \right) $
The interval is given as $ \left( {0,2\pi } \right) $ . So the solution will be
 \[
\Rightarrow x = 0,\pi - 0,\pi + 0 \\
\Rightarrow x = 0,\pi \;
 \] ------(2)
Now dividing the equation(1) with $ \sin x $ , we have
 \[
  \dfrac{1}{{\sin x}} \times \sin x\left( {\sin x + 1} \right) = 0 \times \dfrac{1}{{\sin x}} \\
  \left( {\sin x + 1} \right) = 0 \\
  \sin x = - 1 \\
\Rightarrow x = {\sin ^{ - 1}}\left( { - 1} \right) \;
 \]
The value of $ x $ is an angle having sine value is equal to -1
The interval is given as $ \left( {0,2\pi } \right) $ and we know the sine function is negative in the third quadrant . So the solution will be
 $ x = \dfrac{{3\pi }}{2} $ -------(3)
From equation (2) and (3) we can conclude
 $ x = 0,\dfrac{{3\pi }}{2},\pi $
Therefore the solution of the given trigonometric equation is $ x = 0,\dfrac{{3\pi }}{2},\pi $ .
So, the correct answer is “ $ x = 0,\dfrac{{3\pi }}{2},\pi $ ”.

Note: 1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2. Period of sine function is $ 2\pi $ .
3. The domain of sine function is in the interval $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ and the range is in the interval $ \left[ { - 1,1} \right] $ .