Question

How do you solve $2{{\sin }^{2}}x+\sin x=0$ from $[0,2\pi ]$?

Answer 1

Hint: We will solve this question by considering the expression as a polynomial equation and factorize the equation and solve for the value of $x$ which will get by taking the principal value of $\sin x$ and then write the final solution which lies in the interval of $[0,2\pi ]$.

Complete step-by-step solution:
We have the given expression as $2{{\sin }^{2}}x+\sin x=0$
We can see that the term $\sin x$ is common in both the terms therefore, we can take it out as common and write the expression as:
$\Rightarrow \sin x(2\sin x+1)=0$
We know the property that when $ab=0$, either $a=0$or $b=0$therefore, we can write the expression as:
$\sin x=0\to (1)$
$2\sin x+1=0\to (2)$
Now consider equation $(1)$,
$\sin x=0$
We know $\sin x=0$ when $x=0$, $x=\pi $or $x=2\pi $.
Now consider equation $(2)$,
$2\sin x+1=0$
On transferring $1$ from the left-hand side of the equation to the right-hand side, we get:
$2\sin x=-1$
On dividing both the sides by $2$, we get:
$\sin x=-\dfrac{1}{2}$
Now the principal value of $\sin x=-\dfrac{1}{2}$ is $-\dfrac{\pi }{6}$.
Now since sine is positive in the first and second quadrant, we will subtract the principal value to get the solution in the second quadrant.
Therefore,
$\pi - \left( \dfrac{- \pi }{6} \right)$
On simplifying we get$\dfrac{7\pi }{6}$.
Therefore, the solution set is $0,\pi ,2\pi ,\dfrac{7\pi }{6}$ which lies in the range $[0,2\pi ]$.

Note: The trigonometric table should be remembered to get the appropriate value of the angle, the angle can be written as degrees or radians.
It is to be remembered which trigonometric functions are positive and negative in what quadrants to get the sign of the angle.
We have used the property of coterminal angles which means that the angles which have the same sides in the triangle.
The general form of writing coterminal angles is $\theta \pm k(2\pi )$ where $\theta $ is the initial angle, $k$ is a constant multiplied by $2\pi $ because whenever angle is multiplied by $2\pi $ it has the same sides as the initial angle.