Conditional Trigonometric Equations, Helpo!!

**nee** · August 24th 2008, 11:13 AM

Solve the equation for exact solutions over the interval 0,2∏

tan(squared)x + 3 = 0

**Chris L T521** · August 24th 2008, 11:16 AM

Originally Posted by nee

Solve the equation for exact solutions over the interval 0,2∏

tan(squared)x + 3 = 0

$\tan^2x+3=0$

There is no such x value that causes this statement to be true.

Maybe you mean $\tan^2x{\color{red}-}3=0$ ??

--Chris

**mathgeek777** · August 24th 2008, 11:17 AM

Originally Posted by nee

Solve the equation for exact solutions over the interval 0,2∏

tan(squared)x - 3 = 0

The one you gave doesn't exist so I'm assuming you made a typo.

Add 3, then take the square root of both sides. Look on the unit circle to see which angles correspond to $tan = \sqrt{3}$ over the interval $[0,2\pi)$ , which is pretty much the entire unit circle. Those points will be your solutions.

**nee** · August 24th 2008, 11:20 AM

That's the question as it's written in the text, and the answer given is 0 with a diagonal line through it, whatever that means.

**Moo** · August 24th 2008, 11:24 AM

Originally Posted by nee

That's the question as it's written in the text, and the answer given is 0 with a diagonal line through it, whatever that means.

LoL

$\emptyset$ . This means that there is no solution. More precisely $\{\emptyset\}$ stands for "the set of the solutions is empty"

**nee** · August 24th 2008, 11:27 AM

Oh ok. So what kind of work I'm supposed to show to demonstrate how I came to that conclusion?

**Chris L T521** · August 24th 2008, 11:29 AM

Originally Posted by nee

Oh ok. So what kind of work I'm supposed to show to demonstrate how I came to that conclusion?

Note that you can't have $\tan\vartheta=\pm\sqrt{3}i$

Since the tangent would be complex, there wouldn't be a possible solution for x. So our answer would just be an empty set, $\left\{\emptyset\right\}$

**nee** · August 24th 2008, 11:34 AM

Why can't you have

? When I punch it into my calculator I get a number.

**mathgeek777** · August 24th 2008, 11:44 AM

Originally Posted by nee

Why can't you have

? When I punch it into my calculator I get a number.

An imaginary number isn't on the x-axis. In order for that number to be a solution to the equation, the equation must intersect the x-axis at that point. Since that point doesn't exist on the x-axis when y = 0, then it is not a solution of the equation.

**Chris L T521** · August 24th 2008, 11:45 AM

Originally Posted by nee

Why can't you have

? When I punch it into my calculator I get a number.

Maybe I should have said this: You will not find an appropriate value for $\vartheta$ that falls in the domain $0\leq\vartheta\leq 2\pi$ . By the way, my calculator tells me that $\vartheta=\pm\frac{\pi}{2}\mp\frac{\ln(2-\sqrt{3})}{2}i$ which apparently seems like it won't fall in the interval $0\leq\vartheta\leq 2\pi$ .

--Chris

**nee** · August 24th 2008, 11:45 AM

Thanks!

**nee** · August 24th 2008, 11:47 AM

Thanks!

nee

**flyingsquirrel** · August 24th 2008, 12:12 PM

Hi,

Originally Posted by Moo

More precisely $\{\emptyset\}$ stands for "the set of the solutions is empty"

Originally Posted by Chris L T521

So our answer would just be an empty set, $\left\{\emptyset\right\}$

$\{\emptyset\}$ is the set whose only member is the set $\emptyset$ . Saying that the set of the solutions of $\tan^2x+3=0$ is $\{\emptyset\}$ means that for $x = \emptyset$ the equation is satisfied. In other words, $\tan^2\emptyset +3=0$

. I guess we agree that this doesn't make sense at all. To say that the set of the solutions is empty, simply get rid of the two braces : the set of the solutions of $\tan^2x+3=0$ is $\emptyset$ .

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