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Question: How do you find all solutions for \(5\sqrt{3}\tan x+3=8\sqrt{3}\tan x\)?...

How do you find all solutions for 53tanx+3=83tanx5\sqrt{3}\tan x+3=8\sqrt{3}\tan x?

Explanation

Solution

The given equation in the above question is a simple trigonometric equation which can be solved easily using the basic algebraic manipulations. First of all, we need to separate the variable terms on the LHS and the constant term on the RHS. For this, we first have to subtract 83tanx8\sqrt{3}\tan x from both the sides of the given equation. After this, we have to divide the obtained equation by the coefficient of tanx\tan x so that we will obtain the equation of the form tanx=tanα\tan x=\tan \alpha . The solution to this equation can be written by using the general solution which is given as x=nπ±αx=n\pi \pm \alpha , where n is an integer.

Complete step by step solution:
The trigonometric equation given in the above question is
53tanx+3=83tanx\Rightarrow 5\sqrt{3}\tan x+3=8\sqrt{3}\tan x
We subtract 83tanx8\sqrt{3}\tan x from both the sides of the above equation to get

& \Rightarrow 5\sqrt{3}\tan x+3-8\sqrt{3}\tan x=8\sqrt{3}\tan x-8\sqrt{3}\tan x \\\ & \Rightarrow 3+5\sqrt{3}\tan x-8\sqrt{3}\tan x=0 \\\ & \Rightarrow 3-3\sqrt{3}\tan x=0 \\\ \end{aligned}$$ Now, we add $$3\sqrt{3}\tan x$$ on both the sides of the above equation to get $$\begin{aligned} & \Rightarrow 3-3\sqrt{3}\tan x+3\sqrt{3}\tan x=0+3\sqrt{3}\tan x \\\ & \Rightarrow 3=3\sqrt{3}\tan x \\\ & \Rightarrow 3\sqrt{3}\tan x=3 \\\ \end{aligned}$$ Now, we divide both the sides of the above equation by $$3\sqrt{3}$$ to get $$\begin{aligned} & \Rightarrow \dfrac{3\sqrt{3}\tan x}{3\sqrt{3}}=\dfrac{3}{3\sqrt{3}} \\\ & \Rightarrow \tan x=\dfrac{1}{\sqrt{3}} \\\ \end{aligned}$$ We know that $\tan \left( \dfrac{\pi }{6} \right)=\dfrac{1}{\sqrt{3}}$. Therefore, we can write the RHS of the above equation as $\Rightarrow \tan x=\tan \left( \dfrac{\pi }{6} \right)......\left( i \right)$ The above equation is of the form $\Rightarrow \tan x=\tan \alpha $ We know that the general solution of the above equation is given as $\Rightarrow x=n\pi \pm \alpha $ Therefore, we substitute $\alpha =\dfrac{\pi }{6}$ to get the solution of the equation (i) as $\Rightarrow x=n\pi \pm \dfrac{\pi }{6}$ **Hence, the solutions of the given equation are given by $x=n\pi \pm \dfrac{\pi }{6}$, where n is any integer.** **Note:** The solutions can be obtained by substituting different integral values of n in the solution $x=n\pi \pm \dfrac{\pi }{6}$. We must note that we cannot write a single principle solution of the equation obtained in the above solution as $\tan x=\tan \left( \dfrac{\pi }{6} \right)$. A trigonometric equation always has infinite solutions. We must remember the general solutions of the general trigonometric equations $\cos x=\cos \alpha $, $\sin x=\sin \alpha $, and $\tan x=\tan \alpha $.