Question

How do you solve $2\sin x\cos x + \cos x = 0$ from $0$ to $2 \pi$ ?

Answer 1

As to solve this question. We should know about trigonometric identity.
Trigonometric: the relation between lengths of side to the angle. The relationship is fixed. So, the value of the trigonometric function is fixed for a given angle. The value of a function depends only upon the plan at which it is considered.

Complete step by step solution:
As given in question $2\sin x\cos x + \cos x = 0$ .
In $2\sin x\cos x + \cos x = 0$ , we take $\cos x$ as common. We get,
$\Rightarrow \cos x(2\sin x + 1) = 0$
$\Rightarrow \cos x = 0$
$\Rightarrow (2\sin x + 1) = 0$
For,
$\Rightarrow \cos x = 0$
Whose solution in domain $[0,2\pi ]$ is \left\\{ {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right\\}
$\Rightarrow 2\sin x + 1 = 0$
$\Rightarrow \sin x = - \dfrac{1}{2}$
Whose solution in domain $[0,2\pi ]$ is \left\\{ {\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}} \right\\}
Adding a different solution set for both of them. We get,

Hence, solution set is \left\\{ {\dfrac{\pi }{2},\dfrac{{7\pi }}{6},\dfrac{{3\pi }}{2},\dfrac{{11\pi }}{6}} \right\\}

Note: Trigonometric identity: It is equalities that involve trigonometric function and true for every value of the occurring variable for which both sides of the equality are defined.
Some trigonometric identity:
${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\cos (2x) = {\cos ^2}(x) - {\sin ^2}(x)$
The inverse functions are partial inverse functions for the trigonometric function. For example the inverse for the sine, known as the inverse sine $({\sin ^{ - 1}})$ or arcsine (arcsin or asin), satisfy;
$\sin (\arcsin x) = x$ for $\left| x \right| \leqslant 1$
$\sin (\arcsin x) = x$ for $\left| x \right| \leqslant \dfrac{\pi }{2}$