How do you solve \[2\sin x\cos x + \sin x = 0\] in the interval [0,360] ?

Hint: The above question is based on the concept of trigonometric functions. The main approach towards solving the equation is by splitting it and equating the terms with zero. The terms which are actually trigonometric functions on equating with zero will give the solutions which lie on the graph.

Complete step-by-step answer:
Trigonometric function means the function of the angle between the two sides. It tells us the relation between the angles and sides of the right-angle triangle.
From the above given equation we need to find out the solution of the above given equation. The possible values of unknown which satisfy the equation are called solution of the equation. When we try to solve a trigonometric equation, we try to find out all sets of values of
 \[x\] , which satisfy the given equation.
Now we need to first simplify the above equation,
 \[2\sin x\cos x + \sin x = 0\]
We can take the sine trigonometric function common in the above equation,
 \[\sin x\left( {2\cos x + 1} \right) = 0\]
Now by equating each term of trigonometric functions equal to zero we get two solutions.
 \[\sin x = 0\] or \[2\cos x + 1 = 0\]
Now by isolating cosine function on left hand side,
 \[\sin x = 0\] or \[\cos x = - \dfrac{1}{2}\]
So, considering the first term \[\sin x = 0\] .So sine function is zero when
 \[x = 0,\pi ,2\pi \]
And for cosine function to get the above fraction when
 \[x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}\]
Therefore we get the above solutions.
So, the correct answer is “ \[x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}\] and \[x = 0,\pi ,2\pi \] ”.

Note: An important thing to note is that the value 0 and the fraction \[ - \dfrac{1}{2}\] are the points present on the y -axis. And the values of x i.e., the angle is on the x-axis. where \[x = n\pi \] , where n = 0, ±1, ±2..We call it a general solution of the trigonometric equation sin θ = 0, because for all values of n, this solution satisfies the given equation.