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Question

Question: How do you solve \(\tan (x) = \dfrac{1}{2}\)?...

How do you solve tan(x)=12\tan (x) = \dfrac{1}{2}?

Explanation

Solution

To solve the given question means to find the value of xx. In order to find xx here, we will use inverse trigonometric functions. The inverse trigonometric functions are the inverse functions of the trigonometric functions which are used to obtain an angle from any of the angle’s trigonometric ratios. Here we will be using an inverse tan function to find the solution.

Complete step by step solution:
The trigonometric equation is tan(x)=12\tan (x) = \dfrac{1}{2}.
On applying an inverse tan function, we get
tan1(tanx)=tan1(12)\Rightarrow {\tan ^{ - 1}}(\tan x) = {\tan ^{ - 1}}(\dfrac{1}{2})
x=tan1(12)\Rightarrow x = {\tan ^{ - 1}}(\dfrac{1}{2})
x=26.565o\Rightarrow x = {26.565^o}

Hence the solution of tan(x)=12\tan (x) = \dfrac{1}{2} is x=26.565ox = {26.565^o}.

Note: The domain of the inverse tan function is (,)( - \infty ,\infty )which means domain contains all the real numbers and the range is (π2,π2)( - \dfrac{\pi }{2},\dfrac{\pi }{2}) which means range contains all the angles between π2 - \dfrac{\pi }{2} and π2\dfrac{\pi }{2} but not π2 - \dfrac{\pi }{2} and π2\dfrac{\pi }{2}. In the question given above, when we used tan1{\tan ^{ - 1}}, the domain for it was tan(x)\tan (x). The question itself states that tan(x)=12\tan (x) = \dfrac{1}{2}, hence it passed the domain condition for the inverse tan function. Also tan1(tanx)=x{\tan ^{ - 1}}(\tan x) = x only when x(π2,π2)x \in ( - \dfrac{\pi }{2},\dfrac{\pi }{2}). Had xxnot assumed and lied between (π2,π2)( - \dfrac{\pi }{2},\dfrac{\pi }{2}), we would not have been able to use tan1(tanx)=x{\tan ^{ - 1}}(\tan x) = x.
One more thing that should also be kept in mind while using inverse trigonometric function (specifically inverse tan function for the context given) is that the expressiontan1(x){\tan ^{ - 1}}(x) is not equal to 1tan(x)\dfrac{1}{{\tan (x)}}, which means 1 - 1 is not an exponent of tan(x)\tan (x) here.