Question

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

Answer 1

Hint: From the question given we have to solve the $2\cos x-1=0$. To solve this question first we have to shift the numerical values from left hand side to the right-hand side and then as we know that the general solutions of $\cos x=\cos \theta ,x=2n\pi \pm \theta ,n\in i$, from that we can find the solutions of the given equation.

Complete step by step solution:
From the question given we have to solve the given equation that is,
$\Rightarrow 2\cos x-1=0$
First, we have to shift the $-1$ from left hand side to the right-hand side
By shifting $-1$ from left hand side to the right-hand side, we will get,
$\Rightarrow 2\cos x=1$
First, we have to shift the $2$ from left hand side to the right-hand side
By shifting $2$ from left hand side to the right-hand side, we will get,
$\Rightarrow \cos x=\dfrac{1}{2}$
As we know that the value of
$\Rightarrow \cos \dfrac{\pi }{3}=\dfrac{1}{2}$
From this we will get
$\Rightarrow \cos x=\cos \dfrac{\pi }{3}$
As we know that the general solutions of
$\Rightarrow \cos x=\cos \theta ,x=2n\pi \pm \theta ,n\in i$
Where n belongs to integer,
From this we will get that the solutions of “x” are,
$x=2n\pi \pm \dfrac{\pi }{3},n\in i$
Therefore, the solutions of $2\cos x-1=0$ is $x=2n\pi \pm \dfrac{\pi }{3},n\in i$

Note: Students should recall the formulas of trigonometry before doing this problem, students should not be confused and write $\cos \dfrac{\pi }{6}=\dfrac{1}{2}$ instead of $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$, the whole solution will be wrong.