Question

How do you find all solutions of the equation in the interval \[\left[ {0,\left. {2\pi } \right)} \right.\] given \[\sqrt 3 \tan 3x = 0\]?

Answer 1

Hint: Here, we will solve the given Trigonometric equation by using the inverse trigonometric function and basic mathematical operation. Then we will use the given interval to find the solutions to the Trigonometric Equation. A trigonometric equation is defined as an equation involving the trigonometric ratios.

Formula Used:
Trigonometric Ratio: \[\tan n\pi = 0\]

Complete Step by Step Solution:
We are given with an equation \[\sqrt 3 \tan 3x = 0\]
Now, dividing by \[\sqrt 3 \] on both the sides, we get
\[ \Rightarrow \dfrac{{\sqrt 3 \tan 3x}}{{\sqrt 3 }} = \dfrac{0}{{\sqrt 3 }}\]
Thus, we get
\[ \Rightarrow \tan 3x = 0\]
Taking the inverse of a tangent on both sides, we get
\[ \Rightarrow 3x = {\tan ^{ - 1}}0\]
We know that \[\tan n\pi = 0\] , so we get \[{\tan ^{ - 1}}\left( 0 \right) = n\pi \] .
Thus, we get
\[ \Rightarrow 3x = n\pi \]
Here \[n = 0,1.....\]
Since we are given that the solutions of the equations should lie in the interval \[\left[ {0,\left. {2\pi } \right)} \right.\], so we get
\[ \Rightarrow 3x = 0,\pi \]
By rewriting the equation, we get
\[ \Rightarrow x = \dfrac{0}{3},\dfrac{\pi }{3}\]
\[ \Rightarrow x = 0,\dfrac{\pi }{3}\]

Therefore, the solutions of the equation \[\sqrt 3 \tan 3x = 0\] in the interval \[\left[ {0,\left. {2\pi } \right)} \right.\] is 0 and \[\dfrac{\pi }{3}\].

Note:
We know that Trigonometric identity is an equation that is always true for all the variables. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right-angled triangle with respect to any of its acute angle. We know that the inverse trigonometric function is used to find the missing angles in a right-angled triangle. We should know that an Open interval is an interval that does not include the endpoints and is denoted by () whereas a Closed interval is an interval that includes the endpoints and is denoted by [].