Question

How do you solve the equation $\tan x=4$ in the domain $\left[ 0,2\pi \right]$ ?

Answer 1

Now we are given with a trigonometric equation $\tan x=4$ . Now we will apply the function ${{\tan }^{-1}}$ on both sides. Now we know that ${{f}^{-1}}\left( f\left( x \right) \right)=x$ and we can find the value of ${{\tan }^{-1}}\left( 4 \right)$ . Now we also know that $\tan \left( x+\pi \right)=\tan \left( x \right)$ Hence we can find all the solution in the domain $\left[ 0,2\pi \right]$

Complete step by step solution:
Now let us first understand the concept of inverse functions.
Now we know that functions are maps from one set to another.
Let us say we have a function defined as $f:X\to Y$ .
Now hence we have f(x) = y.
Now we can say that with the function f we have y corresponding to x.
Now similarly inverse function is a function which reverses the value of f.
Hence if we have f(x) = y, then we have ${{f}^{-1}}\left( y \right)=x$ where ${{f}^{-1}}$ is the inverse function of x.
Now note that the inverse function is defined as ${{f}^{-1}}:Y\to X$ such that ${{f}^{-1}}\left( y \right)=x$ whenever f(x) = y.
Now consider ${{f}^{-1}}\left( f\left( x \right) \right)$ , Let f(x) = y, then we have ${{f}^{-1}}\left( y \right)=x$ .
Hence we have ${{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( y \right)=x$
Similarly we have inverse trigonometric functions corresponding to each trigonometric ratio.
Now consider the equation $\tan x=4$ .
Applying ${{\tan }^{-1}}$ on both sides we get,
${{\tan }^{-1}}\left( \tan x \right)={{\tan }^{-1}}4$
Now we know that ${{\tan }^{-1}}\left( \tan x \right)=x$
Hence we have $x={{\tan }^{-1}}4$
Taking the value of ${{\tan }^{-1}}$ from trigonometric sheet we get,
$x=1.326$ .
Note that the value is in radians.
Now we know that the angle 1.326 is in the first quadrant. Also we know that $\tan \left( x+\pi \right)=\tan x$
Hence there will also be one solution in third quadrant given by $x+\pi =1.326+3.14=4.465$
Hence the solutions to the equation are 1.326 and 4.465.

Note: Now note that inverse function is just denoted by ${{f}^{-1}}$ and is not equal to $\dfrac{1}{f}$ . Also note that not all functions will have an inverse function. A function must be bijective which means it must be one one and onto for the inverse function to exist.