Question

How do you solve $2\cos x+1=0$?

Answer 1

Hint: The given trigonometric equation contains only one function which is $\cos x$. Hence we don’t have to simplify the equation any further.
The cosine function is a ratio of the adjacent side to the hypotenuse. If we know the cosine value of some important angles, it will be useful in finding the adjacent side and hypotenuse, given that any one of these sides is known.

Complete Step by Step Solution:
The given trigonometric equation is $2\cos x+1=0$. We can rearrange this equation as
$\Rightarrow 2\cos x=-1$
$\Rightarrow \cos x=\dfrac{-1}{2}$
Hence the variable x will be equal to
$\Rightarrow x={{\cos }^{-1}}\left( -\dfrac{1}{2} \right)$
We know that the cosine function will be negative in quadrants 2 and 3. This is because of the negative value of the adjacent side in these quadrants.

In the second quadrant, the cosine value will be equal to $-\dfrac{1}{2}$ at ${{120}^{{}^\circ }}$ or $\dfrac{2\pi }{3}$ , and in the second quadrant, the cosine value will be $-\dfrac{1}{2}$ at ${{240}^{{}^\circ }}$ or $\dfrac{4\pi }{3}$ as shown in the above figure.
And the cosine function is periodic with a period equal to $2\pi $.

Therefore the variable x will be given by the following equation.
$\Rightarrow x=\dfrac{2\pi }{3}+2\pi n,\dfrac{4\pi }{3}+2\pi n$ where n is an integer.

Note:
It is important to know the sign of the trigonometric ratios in all four quadrants. This will help us to sort out the position of the point in the coordinate system. In addition to that, we should consider the periodicity property of these trigonometric functions and we should include the period of the particular function in the final result.