# Math Help - Solving Trigonometric Equation (Dividing by cos x)

1. ## Solving Trigonometric Equation (Dividing by cos x)

Hi all ---

I'm trying to solve this question - and I've included the solution --

Question 1 --- I can't understand why you can divide by $\cos x$.

Because $0 \leq x < 2 \pi$, you can have $\cos\left(\pi/2)\right = 0 = \cos\left(3\pi/2)\right$, right?

And we don't know what the x-values are. We haven't solved the equation yet?

Question 2 --- Also, how'd you know to divide by $\cos x$? Why not $\sin x$?

Thanks all.

2. ## Re: Solving Trigonometric Equation (Dividing by cos x)

hi

A1 if you check x=pi/2 + k*pi you will see that those aren't the answers to the equation,so you can divide by cos(x) without the fear of dividing by zero.

A2 if you divide by sin(x) you will get an equation in terms of cot(x),so an equation with terms in tan(x) might be better for solving.anyway if you deal with cot(x) as well as with tan(x) than you can also divide by sin(x).

PS never,i mean never divide by zero.you can make that mistake only once.

3. ## Re: Solving Trigonometric Equation (Dividing by cos x)

Hi anonimnystefy ---

I understand what you wrote perfectly. But how should I've seen that I had to divide by $\cos x$? Of course - after you look at the solution - it's "obvious".

But when I was doing the problem. I was very unsure if I should divide by $\cos x$. How would I know to check that the values that give $\cos x = 0$ aren't actually the solutions to the given equation?

4. ## Re: Solving Trigonometric Equation (Dividing by cos x)

Originally Posted by mathminor827
Hi anonimnystefy ---

I understand what you wrote perfectly. But how should I've seen that I had to divide by $\cos x$? Of course - after you look at the solution - it's "obvious".

But when I was doing the problem. I was very unsure if I should divide by $\cos x$. How would I know to check that the values that give $\cos x = 0$ aren't actually the solutions to the given equation?
You don't need to divide by $\displaystyle \cos{x}$ in the beginning, if you recognise that $\displaystyle \sin^2{x} + 3\sin{x}\cos{x} + 2\cos^2{x} = (\sin{x} + \cos{x})(\sin{x} + 2\cos{x})$

5. ## Re: Solving Trigonometric Equation (Dividing by cos x)

Originally Posted by Prove It
You don't need to divide by $\displaystyle \cos{x}$ in the beginning, if you recognise that $\displaystyle \sin^2{x} + 3\sin{x}\cos{x} + 2\cos^2{x} = (\sin{x} + \cos{x})(\sin{x} + 2\cos{x})$
Hi Prove It ---

But $\sin^2{x} + 3\sin{x}\cos{x} + 2\cos^2{x} = (\sin{x} + \cos{x})(\sin{x} + 2\cos{x}) = 0$ means ---

$\sin{x} + \cos{x} = 0$ or $\sin{x} + 2\cos{x} = 0$

Do these 2 equations really make it easier to solve the problem?

I don't think you can find $x$ directly? I mean, from the first, I get ---
$\sin{x} = -\cos{x}$ and after I try some numbers out, I get as I should --- $x = 3/4 \pi$?

Or can someone explain how someone doing this problem for the first time can "see" to divide both sides by $\cos x$ in the given solution?

Thanks again ---

6. ## Re: Solving Trigonometric Equation (Dividing by cos x)

Why to divide by $\cos^2(x)$ is just something you have to try and with some experience there're some people who can predict that but just try some possibilities. If you exercice a lot then you will be better to see things like that.

You can solve this equation, for example:
$\sin(x)=-\cos(x)$
You can divide by $\cos(x)$ because $\frac{\pi}{2}+2k\pi$ is no solution therefore:
$\frac{\sin(x)}{\cos(x)}=-1$
...

Try the same for the second one.