# Solution of Trigonometric equations

If we were to solve the trigonometric equation $$\sqrt{13-18\tan x} = 6\tan x-3$$ by squaring both the sides, we would get two roots; $\tan x = \frac{2}{3}$ and $\tan x=-\frac{1}{6}$

2/3 is okay, but when we substitute -1/6 in the equation and simplify it, it becomes $\sqrt{16}$ = $-4$

The solution to the question says that this must be rejected, as $\sqrt{16}$ = $|4|$, which cannot be equal to $-4$.

Why is this done? Why can't the square root of $16$ be equated to $-4$?

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Square root is a function and has only one value which is the positive one. Square root is the function $y=\sqrt {x}$. Don't confuse it with $y=x^{2}$, which is a relation.
Hence $\sqrt {16}=4$ and not $-4$.

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Thanks for that perspective. I looked around a bit more, and found that it's a very satisfying answer. Thanks once again! –  Dhananjay Gupta Jan 6 '14 at 16:28
please upvote and accept the answer if you liked it.. –  Apurv Jan 6 '14 at 16:29

What you need to know is that $$\sqrt{13-18\tan x}=6\tan x-3$$ is not the same as $$13-18\tan x=(6\tan x-3)^2.$$ These are different equations.

Notice that $$\sqrt{13-18\tan x}=6\tan x-3$$ $$\iff 13-18\tan x=(6\tan x-3)^2\ \ \text{and}\ \ 6\tan x-3\ge0$$ $$\iff 13-18\tan x=(6\tan x-3)^2\ \ \text{and}\ \ \tan x\ge\frac 12.$$

So, we know that $\tan x=-1/6$ is not a solution of the equation at the top.

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I see. But i'm still unable to understand why you had to define RHS as a positive value when you squared both sides. Why does the fact that we're taught in elementary math, that sqrt 16 = 4 or -4, invalid here? Is it because the notation $\sqrt{16}$ can only be used for 4, not -4? –  Dhananjay Gupta Jan 6 '14 at 16:15
Note that $\sqrt{16}$ is positive. It is already positive. So, $\sqrt{13-18\tan x}\ge0$. For example, if you have an equation $\sqrt x=-2,$ do you think you can solve it? –  mathlove Jan 6 '14 at 16:18
Oh okay. So in the original equation, LHS already is positive, which is why, upon solving, RHS cannot be negative. I understand. Thanks a lot for your help! –  Dhananjay Gupta Jan 6 '14 at 16:23
You are welcome! –  mathlove Jan 6 '14 at 16:35

The real problem is that squaring both sides of an equation may introduce false solutions to the original equation. For example, if you start with $$x = -1$$ and square both sides, you get $$x^2 = 1$$ which has solutions $x=1$ and $x=-1$. Obviously only one of these is a solution to the original equation. You must check your solutions in the original equation if you square and solve that way.

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Why can't the square root of 16 be equated to -4?

It can be... But in a different context... Later on, when you'll learn about complex numbers, they'll teach you that each radical or order n returns n distinct complex values. E.g., $\sqrt[4]1=\{\pm1,\pm i\}$. In other words, a radical isn't really a function, but a relation, since a function, by definition, returns exactly one value. If, however, you do want to transform it into one, then you have to make sure that it fulfills this demand, and the only way to do this is to restrict its value domain. And since we are only working with real numbers (for now), and since people are more familiar with positive numbers than with negative ones, then, by convention, we choose this domain to be restricted to the positive reals, or $\mathbb{R}^+$.

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Yes, i know about complex numbers, and the complex roots of unity, but what i was confused about was the more fundamental fact that my teachers taught me about when I was younger. I now see that my understanding was correct, but because of a lack of proper knowledge of relations and functions, I was getting confused. Thanks for the help! –  Dhananjay Gupta Jan 10 '14 at 17:14