Question

How do you find all solutions (in the range $0 < x < \pi $) of the equation $\cos 2x = \dfrac{{ - \sqrt 3 }}{2}?$

Answer 1

Hint:First find the general solution of the given trigonometric equation and then shrink that general solution into the given range for the required particular solution.

Complete step by step solution:
Given trigonometric equation $\cos 2x = \dfrac{{ - \sqrt 3 }}{2}$, we have to find its solution in the range $0 < x < \pi $
In order to find the solutions for $x$ in the range $0 < x < \pi $ we will first find the general solution of the equation $\cos 2x = \dfrac{{ - \sqrt 3 }}{2}$
The principal values for $\cos \theta = \dfrac{{ - \sqrt 3 }}{2}$ are follows
$\theta = \dfrac{{ - 5\pi }}{6}\;{\text{and}}\;\dfrac{{5\pi }}{6}$
This is the principal solution, to find the general solution we have to add the period of the cosine function which is $2\pi $
$
\Rightarrow \theta = \dfrac{{ - 5\pi }}{6}\;{\text{and}}\;\dfrac{{5\pi }}{6} \\
\Rightarrow \theta = 2n\pi - \dfrac{{5\pi }}{6}\;{\text{and}}\;2n\pi + \dfrac{{5\pi
}}{6},\;{\text{where}}\;n \in I \\
\Rightarrow \theta = 2n\pi \pm \dfrac{{5\pi }}{6},\;{\text{where}}\;n \in I \\
$
So now, for $\cos 2x = \dfrac{{ - \sqrt 3 }}{2}$, the general solution will be written as
$
\Rightarrow 2x = 2n\pi \pm \dfrac{{5\pi }}{6},\;{\text{where}}\;n \in I \\
\Rightarrow x = \dfrac{{2n\pi \pm \dfrac{{5\pi }}{6}}}{2},\;{\text{where}}\;n \in I \\
\Rightarrow x = n\pi \pm \dfrac{{5\pi }}{{12}},\;{\text{where}}\;n \in I \\
$
So we got the general solution for $\cos 2x = \dfrac{{ - \sqrt 3 }}{2}$, which is $x = n\pi \pm
\dfrac{{5\pi }}{{12}},\;{\text{where}}\;n \in I$
Now we will choose the value of $n$ so that the value of $x$ lies in the range $\left[ {0,\;\pi } \right]$
For this we will do the following
$
0 < x < \pi \\
0 < n\pi \pm \dfrac{{5\pi }}{{12}} < \pi ,\;{\text{where}}\;n \in I \\
$
Now we will take both cases, $\dfrac{{5\pi }}{{12}}\;{\text{and}}\;\dfrac{{ - 5\pi }}{{12}}$
For $\dfrac{{5\pi }}{{12}}$,
$
0 < n\pi + \dfrac{{5\pi }}{{12}} < \pi ,\;{\text{where}}\;n \in I \\
0 - \dfrac{{5\pi }}{{12}} < n\pi < \pi - \dfrac{{5\pi }}{{12}},\;{\text{where}}\;n \in I \\
- \dfrac{{5\pi }}{{12}} < n\pi < \dfrac{{7\pi }}{{12}},\;{\text{where}}\;n \in I \\
$
Dividing $\pi $ from all, we will get
$
- \dfrac{{5\pi }}{{12\pi }} < \dfrac{{n\pi }}{\pi } < \dfrac{{7\pi }}{{12\pi }},\;{\text{where}}\;n \in I
\\
- \dfrac{5}{{12}} < n < \dfrac{7}{{12}},\;{\text{where}}\;n \in I \\
$
$\therefore $ we get the value of $n = 0$ in this case, putting this in the general solution
$
\Rightarrow x = n\pi \pm \dfrac{{5\pi }}{{12}},\;{\text{where}}\;n \in I \\
\Rightarrow x = 0 \times \pi \pm \dfrac{{5\pi }}{{12}} \\
\Rightarrow x = \pm \dfrac{{5\pi }}{{12}} \\
$
But $ - \dfrac{{5\pi }}{{12}}$ doesn’t lie in the range $\left[ {0,\;\pi } \right]$
$\therefore x = \dfrac{{5\pi }}{{12}}$ is the solution in this case
For $\dfrac{{ - 5\pi }}{{12}}$ ,
$
0 < n\pi - \dfrac{{5\pi }}{{12}} < \pi ,\;{\text{where}}\;n \in I \\
0 + \dfrac{{5\pi }}{{12}} < n\pi < \pi + \dfrac{{5\pi }}{{12}},\;{\text{where}}\;n \in I \\
\dfrac{{5\pi }}{{12}} < n\pi < \dfrac{{17\pi }}{{12}},\;{\text{where}}\;n \in I \\
$
Dividing $\pi $ from all, we will get
$
\dfrac{{5\pi }}{{12}} < n\pi < \dfrac{{17\pi }}{{12}},\;{\text{where}}\;n \in I \\
\dfrac{{5\pi }}{{12\pi }} < \dfrac{{n\pi }}{\pi } < \dfrac{{17\pi }}{{12\pi }},\;{\text{where}}\;n \in I
\\
\dfrac{5}{{12}} < n < \dfrac{{17}}{{12}},\;{\text{where}}\;n \in I \\
$
$\therefore $ we get the value of $n = 1$ this time, putting it in general solution
$
\Rightarrow x = n\pi \pm \dfrac{{5\pi }}{{12}},\;{\text{where}}\;n \in I \\
\Rightarrow x = 1 \times \pi \pm \dfrac{{5\pi }}{{12}} \\
\Rightarrow x = \pi \pm \dfrac{{5\pi }}{{12}} \\
\Rightarrow x = \pi + \dfrac{{5\pi }}{{12}}\;and\;\pi - \dfrac{{5\pi }}{{12}} \\
\Rightarrow x = \dfrac{{17\pi }}{{12}}\;{\text{and}}\;\dfrac{{7\pi }}{{12}} \\
$
Here $x = \dfrac{{17\pi }}{{12}}$ is not in the required range
$\therefore x = \dfrac{{7\pi }}{{12}}$ is the solution for this case.
That is the overall solution for $\cos 2x = \dfrac{{ - \sqrt 3 }}{2}$ in range $\left[ {0,\;\pi } \right]$ is $x = \dfrac{{5\pi }}{{12}}\;{\text{and}}\;\dfrac{{7\pi }}{{12}}$

Note: The solution in this problem is called the particular solution of the equation in which some particular values come up from the general solution, satisfying all the conditions given in the problem. So read the conditions in the question (if given) and then answer accordingly.