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Question: How do you solve \(\tan x=5\) ?...

How do you solve tan⁡x=5\tan x=5 ?

Explanation

Solution

To solve the given trigonometric equation i.e. tan⁡x=5\tan x=5, we need to calculate the value of x which we are going to find out by taking tan⁡−1{{\tan }^{-1}} on both the sides of this trigonometric equation. After the tan⁡−1{{\tan }^{-1}} application on both the sides of the equation, the L.H.S of the equation becomes 1 because multiplying a mathematical entity with its inverse we will get 1 and we are multiplying tan⁡−1{{\tan }^{-1}} by tan.

Complete step by step answer:
The trigonometric equation given in the above problem is as follows:
tan⁡x=5\tan x=5
The solution of the above equation is the value of x. To get that value of x, we are going to take tan⁡−1{{\tan }^{-1}} on both the sides we get,
⇒tan⁡−1(tan⁡x)=tan⁡−1(5)\Rightarrow {{\tan }^{-1}}\left( \tan x \right)={{\tan }^{-1}}\left( 5 \right)
We know that the multiplication of a mathematical expression and its inverse is 1 so using this relation in the L.H.S of the above equation we get,
⇒1.x=tan⁡−15 ⇒x=tan⁡−15 \begin{aligned} & \Rightarrow 1.x={{\tan }^{-1}}5 \\\ & \Rightarrow x={{\tan }^{-1}}5 \\\ \end{aligned}
In the above equation, x is the angle in tan such that tangent of x is equal to 5. According to tangent values at different angles, the angle at which tan is 5 is equal to:
78∘69′{{78}^{\circ }}69'
We also know that tan is positive in the first and third quadrant so the general form of the above angle is as follows:
78∘69′+nπ{{78}^{\circ }}69'+n\pi
In the above expression, n can take values such as 0, 1, 2…….n. The above expression gives us the value in two quadrants first and third.
Now, substituting the value of tan⁡−15{{\tan }^{-1}}5 which we have calculated above in x=tan⁡−15x={{\tan }^{-1}}5 we get,
⇒x=78∘69′+nπ\Rightarrow x={{78}^{\circ }}69'+n\pi

Hence, the solution of the above trigonometric equation is 78∘69′+nπ{{78}^{\circ }}69'+n\pi .

Note: The mistake that could be possible in the above problem is that you might forget to write the solution in the third quadrant and have just written the solution of the first quadrant i.e. 78∘69′{{78}^{\circ }}69'. So, make sure you won’t forget to write the solution of the third quadrant also.