Question

How do you solve $2\cos x=\sqrt{3}$ ?

Answer 1

Hint: We have been given a linear equation in the trigonometric function, cosine of x which also consists of a constant term, square root of 3, on the left hand side. Thus, we shall first divide both sides to make the coefficient of cos x equal to 1. In order the value of x, we will compare with the respective value of the cosine function value.

Complete step by step solution:
Given that $2\cos x=\sqrt{3}$.
On dividing both sides by 2 to make the coefficient of cos x equal to 1, we get
$\Rightarrow \cos \left( x \right)=\dfrac{\sqrt{3}}{2}$
We know that $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$ for its principle solution set, that is, in the interval $\left[ 0,2\pi \right]$.
Comparing this value of cosine function with the calculated value from the given equation, we get
$\Rightarrow \cos \left( x \right)=\cos \dfrac{\pi }{6}$
Therefore, we get $x=\dfrac{\pi }{6}$.
Since, our aim is to obtain value when the coefficient of x equal to 1 in order to obtain a well-defined solution for given equation, thus the principle solution for $2\cos x=\sqrt{3}$ is $x=\dfrac{\pi }{6}$.
However, the general solution would be calculated as $\dfrac{\pi }{6}+2n\pi $ where n is any integer.
Therefore, the solution for the given equation $2\cos x=\sqrt{3}$ is $\dfrac{\pi }{6}+2n\pi $ where $n\in \mathbb{Z}$.

Note: We must have prior knowledge of the six basic trigonometric functions, namely, sine, cosine, tangent, cotangent, secant and cosecant to solve such equations as given in this problem. The best method to deeply understand these functions and solve trigonometric equations instantly is by memorizing the graphs of these functions in the principal value set at least which lies in the interval $0\le x\le 2\pi $.