Question

How do you solve $\cos \left( t \right) = - \dfrac{{\sqrt 3 }}{2}$ over the interval $0$ to $2\pi$?

Answer 1

Hint: First, find the values of $t$ satisfying $\cos \left( t \right) = - \dfrac{{\sqrt 3 }}{2}$ using trigonometric properties. Next, find all values of $t$ in the interval $\left[ {0,2\pi } \right]$. Then, we will get all solutions of the given equation in the given interval.
Formula used:
1. $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$
2. $\cos \left( {\pi - x} \right) = - \cos x$
3. $\cos \left( {\pi + x} \right) = - \cos x$

Complete step by step solution:
Given equation: $\cos \left( t \right) = - \dfrac{{\sqrt 3 }}{2}$
We have to find all possible values of $t$ satisfying a given equation over the interval $0$ to $2\pi$.
First, we will find the values of $t$ satisfying $\cos \left( t \right) = - \dfrac{{\sqrt 3 }}{2}$…(i)
So, using the property $\cos \left( {\pi - x} \right) = - \cos x$ and $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$ in equation (i).
$ \Rightarrow \cos t = - \cos \dfrac{\pi }{6}$
$ \Rightarrow \cos t = \cos \left( {\pi - \dfrac{\pi }{6}} \right)$
$ \Rightarrow t = \dfrac{{5\pi }}{6}$
Now, using the property $\cos \left( {\pi + x} \right) = - \cos x$ and $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$ in equation (i).
$ \Rightarrow \cos t = - \cos \dfrac{\pi }{6}$
$ \Rightarrow \cos t = \cos \left( {\pi + \dfrac{\pi }{6}} \right)$
$ \Rightarrow t = \dfrac{{7\pi }}{6}$
Since, the period of the $\cos t$ function is $2\pi $ so values will repeat every $2\pi $ radians in both directions.
$t = \dfrac{{5\pi }}{6} + 2n\pi ,\dfrac{{7\pi }}{6} + 2n\pi $, for any integer $n$.
Now, find the values of $n$ that produce a value within the interval $\left[ {0,2\pi } \right]$.
i.e., find all values of $t$ over the interval $0$ to $2\pi$.
Plug in $0$ for $n$ and simplify to see if the solution is contained in $\left[ {0,2\pi } \right]$.
Since, it is given that $t \in \left[ {0,2\pi } \right]$, hence put $n = 0$ in the general solution.
So, putting $n = 0$ in $t = \dfrac{{5\pi }}{6} + 2n\pi ,\dfrac{{7\pi }}{6} + 2n\pi $, we get
$t = \dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6}$
Thus, $t = \dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6}$ or $t = {150^ \circ },{210^ \circ }$.

Hence, $t = \dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6}$ or $t = {150^ \circ },{210^ \circ }$ are solutions of the given equation over the interval $0$ to $2\pi.$

Note: In above question, we can find the solutions of given equation by plotting the equation, $\cos \left( t \right) = - \dfrac{{\sqrt 3 }}{2}$ on graph paper and determine all solutions which lie in the interval, $\left[ {0,2\pi } \right]$.