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Question: How do you solve \( {\sin ^2}x + \sin x = 0 \) and solve the equation on the interval of \( \left( {...

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

Explanation

Solution

Hint : In order to determine the solution of the above trigonometric equation, pull out sin⁡x\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 (0,2π)\left( {0,2\pi } \right) .

Complete step by step solution:
We are given a trigonometric equation sin⁡2x+sin⁡x=0{\sin ^2}x + \sin x = 0 and we have to find its solution
sin⁡2x+sin⁡x=0{\sin ^2}x + \sin x = 0
Pulling out sin⁡x\sin x as common from the terms, we get
sin⁡x(sin⁡x+1)=0\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\sin x and (sin⁡x+1)\left( {\sin x + 1} \right) one by one
First Dividing both sides of equation with (sin⁡x+1)\left( {\sin x + 1} \right) , we have

1(sin⁡x+1)×sin⁡x(sin⁡x+1)=0×1(sin⁡x+1) sin⁡x=0 x=sin⁡−1(0)   \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 xx is an angle having sine value 0. Since we know sin⁡0=0→0=sin⁡−1(0)\sin 0 = 0 \to 0 = {\sin ^{ - 1}}\left( 0 \right)
The interval is given as (0,2π)\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] $ .