Question

How do you solve $\left| {\sin x} \right| = \dfrac{{\sqrt 3 }}{2}$ in the interval $\left[ {0,{{360}^ \circ }} \right]$?

Answer 1

Hint:
First, find the values of $x$ satisfying $\sin x = \dfrac{{\sqrt 3 }}{2}$ using trigonometric properties.
Next, find the values of $x$ satisfying $\sin x = - \dfrac{{\sqrt 3 }}{2}$ using trigonometric properties. Next, find all values of $x$ in the interval $\left[ {0,{{360}^ \circ }} \right]$. Then, we will get all solutions of the given equation in the given interval.

Formula used:
1) $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
2) $\sin \left( {\pi + x} \right) = - \sin x$
3) $\sin \left( {2\pi - x} \right) = - \sin x$

Complete step by step solution:
Given equation: $\left| {\sin x} \right| = \dfrac{{\sqrt 3 }}{2}$
We have to find all possible values of $x$ satisfying given equation in the interval $\left[ {0,{{360}^ \circ }} \right]$.
First, we will find the values of $x$ satisfying $\sin x = \dfrac{{\sqrt 3 }}{2}$.
So, take the inverse sine of both sides of the equation to extract $x$ from inside the sine.
$x = \arcsin \left( {\dfrac{{\sqrt 3 }}{2}} \right)$
Since, the exact value of $\arcsin \left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{\pi }{3}$.
$ \Rightarrow x = \dfrac{\pi }{3}$
Since, the sine function is positive in the first and second quadrants.
So, to find the second solution, subtract the reference angle from $\pi $ to find the solution in the second quadrant.
$x = \pi - \dfrac{\pi }{3}$
$ \Rightarrow x = \dfrac{{2\pi }}{3}$
Since, the period of the $\sin x$ function is $2\pi $ so values will repeat every $2\pi $ radians in both directions.
$x = \dfrac{\pi }{3} + 2n\pi ,\dfrac{{2\pi }}{3} + 2n\pi $, for any integer $n$.
Now, we will find the values of $x$ satisfying $\sin x = - \dfrac{{\sqrt 3 }}{2}$…(i)
So, using the property $\sin \left( {\pi + x} \right) = - \sin x$ and $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$ in equation (i).
$ \Rightarrow \sin x = - \sin \dfrac{\pi }{3}$
$ \Rightarrow \sin x = \sin \left( {\pi + \dfrac{\pi }{3}} \right)$
$ \Rightarrow x = \dfrac{{4\pi }}{3}$
Now, using the property $\sin \left( {2\pi - x} \right) = - \sin x$ and $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$ in equation (i).
$ \Rightarrow \sin x = - \sin \dfrac{\pi }{3}$
$ \Rightarrow \sin x = \sin \left( {2\pi - \dfrac{\pi }{3}} \right)$
$ \Rightarrow x = \dfrac{{5\pi }}{3}$
Since, the period of the $\sin x$ function is $2\pi $ so values will repeat every $2\pi $ radians in both directions.
$x = \dfrac{{4\pi }}{3} + 2n\pi ,\dfrac{{5\pi }}{3} + 2n\pi $, for any integer $n$.
Now, find all values of $x$ in the interval $\left[ {0,{{360}^ \circ }} \right]$.
Since, it is given that $x \in \left[ {0,{{360}^ \circ }} \right]$, hence put $n = 0$ in the general solution.
So, putting $n = 0$ in $x = \dfrac{\pi }{3} + 2n\pi ,\dfrac{{2\pi }}{3} + 2n\pi $, we get
$x = \dfrac{\pi }{3},\dfrac{{2\pi }}{3}$
Now, putting $n = 0$ in $x = \dfrac{{4\pi }}{3} + 2n\pi ,\dfrac{{5\pi }}{3} + 2n\pi $,we get
$x = \dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3}$
Thus, $x = \dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3}$ or $x = {60^ \circ },{120^ \circ },{240^ \circ },{300^ \circ }$.

Final solution: Hence, $x = \dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3}$ or $x = {60^ \circ },{120^ \circ },{240^ \circ },{300^ \circ }$ are solutions of the given equation in the interval $\left[ {0,{{360}^ \circ }} \right]$.

Note:
In above question, we can find the solutions of given equation by plotting the equation, $\left| {\sin x} \right| = \dfrac{{\sqrt 3 }}{2}$ on graph paper and determine all solutions which lie in the interval, $\left[ {0,{{360}^ \circ }} \right]$.

From the graph paper, we can see that there are four values of $x$ in the interval $\left[ {0,{{360}^ \circ }} \right]$.
So, these will be the solutions of the given equation in the given interval.
Final solution: Hence, $x = \dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{5\pi }}{3}$ or $x = {60^ \circ },{120^ \circ },{240^ \circ },{300^ \circ }$ are solutions of the given equation in the interval $\left[ {0,{{360}^ \circ }} \right]$.