Question

How do you solve $4\tan 3x+5=1$ for $0\le x\le \pi$?

Answer 1

In this question, we are given a trigonometric equation and we need to find the value of x which satisfies the equation between the interval 0 to $\pi$. For this we will rearrange the equation and find one value of tany such that tan3x = tany. After that we will find the general value of x using the formula that if $\tan x=\tan y\Rightarrow x=n\pi \pm y,n\in Z$.
Then we will take the value of x such that we have value of x in the interval $0\le x\le \pi$.

Complete step by step answer:
Here we are given the trigonometric equation as $4\tan 3x+5=1$.
We need to find the value of x satisfying the equation lying between 0 and $\pi$.
For this, let us first rearrange the equation we have $4\tan 3x+5=1$.
Subtracting 5 from both sides of the equation we get $4\tan 3x+5-5=1-5$.
Simplifying we get $4\tan 3x=-4$.
Dividing both sides of the equation by 4 we get $\dfrac{4\tan 3x}{4}=\dfrac{-4}{4}$.
Simplifying we get $\tan 3x=1$.
Now let us try to find some y such that tany = 1. We know that $\tan \dfrac{\pi }{4}=1$ but we need a negative sign with 1. As we know, tan is negative in the second quadrant which can be written as $\pi -\dfrac{\pi }{4}$. So we can say that $\tan \left( \pi -\dfrac{\pi }{4} \right)=-1$.
Simplifying the angle we get $\tan \dfrac{3\pi }{4}=-1$ so we get $\tan 3x=\tan \dfrac{3\pi }{4}$.
We know that $\tan x=\tan y\Rightarrow x=n\pi \pm y,n\in Z$.
So let us use this, we have $\tan 3x=\tan \dfrac{3\pi }{4}$.
So this implies that $3x=n\pi +\dfrac{3\pi }{4},n\in Z$.
Dividing both sides by 3 we get $x=\dfrac{n\pi }{3}+\dfrac{\pi }{4},n\in Z$.
These are the general values of x for this equation. But we require only the values lying between 0 to $\pi$.
So let us put x = 0 we get $x=0+\dfrac{\pi }{4}\Rightarrow x=\dfrac{\pi }{4}$.
As $x=\dfrac{\pi }{4}$ satisfies the condition $0\le x\le \pi$. So one value of x is $\dfrac{\pi }{4}$.
Putting x = 1 we get $x=\dfrac{\pi }{3}+\dfrac{\pi }{4}$.
Simplifying the value of x by taking the LCM as 12 we get $x=\dfrac{7\pi }{12}$.
As $x=\dfrac{7\pi }{12}$ satisfies the condition $0\le x\le \pi$. So one of the values of x is $\dfrac{7\pi }{12}$.
Putting x = 2 we get $x=\dfrac{2\pi }{3}+\dfrac{\pi }{4}$.
Simplifying the value of x by taking the LCM as 12 we get $x=\dfrac{11\pi }{12}$.
As $x=\dfrac{11\pi }{12}$ satisfies the condition $0\le x\le \pi$. So one of the values of x is $\dfrac{11\pi }{12}$.
Any values other than them will not satisfy the condition $0\le x\le \pi$.
So the required value of x are $\dfrac{\pi }{4}$, $\dfrac{7\pi }{12}$, $\dfrac{11\pi }{12}$.

Note:
Students should find all the values of x lying in the given interval. Note that, here Z means the set of integers. Take care of calculation of angles. Students should keep in mind all the trigonometric formulas along with taking care of signs to solve this sum.