# Math Help - [SOLVED] Solving Trigonometric Functions algebraically

1. ## [SOLVED] Solving Trigonometric Functions algebraically

Hello, first time posting.

Had a quick question on a problem I was doing.

The problem:

Find exact solutions

$\cos x = \cot x$

What I've done so far is

$\cos x = \cot x$

$\cos x = (\cos x / \sin x)$

$\sin x = \cos x / \cos x )$

$\sin x = 1$

Now, when $\sin x = 1$ then x is $\pi / 2$. But the answer is also $3 \pi / 2$

But isn't $3 \pi / 2 = -1$?

I'm missing a part of the theory obviously, was just wondering if someone could clarify.

Thank you!

2. but $\cos{\frac{3\pi}{2}}$ = 0

3. When cosx is positive, cotx may be positive or negative.
So square both the sides, you get
cos^x = cot^2x
Cos^2x = 0 or sin^2x = 1 or sinx = (+ or -) 1
Hence x = 90 οr 270 degrees

4. Originally Posted by sleels
Hello, first time posting.

Had a quick question on a problem I was doing.

The problem:

Find exact solutions

$\cos x = \cot x$

What I've done so far is

$\cos x = \cot x$

$\cos x = (\cos x / \sin x)$

$\sin x = \cos x / \cos x )$

$\sin x = 1$

Now, when $\sin x = 1$ then x is $\pi / 2$. But the answer is also $3 \pi / 2$

But isn't $3 \pi / 2 = -1$?

I'm missing a part of the theory obviously, was just wondering if someone could clarify.

Thank you!
$\cos x = \cot x$

$\Rightarrow \cos x = \frac{\cos x}{\sin x}$

$\Rightarrow \sin x \cos x = \cos x$

$\Rightarrow \sin x \cos x - \cos x = 0$

$\Rightarrow \cos x (\sin x - 1) = 0$.

Case 1: $\cos x = 0$.

Case 2: $\sin x - 1 = 0$.

Solve each case over the given domain.

By cancelling the cos x in your solution you fell into the classic trap for young players.

5. Originally Posted by sleels
Hello, first time posting.

Had a quick question on a problem I was doing.

The problem:

Find exact solutions

$\cos x = \cot x$

What I've done so far is

$\cos x = \cot x$

$\cos x = (\cos x / \sin x)$

$\sin x = \cos x / \cos x )$
This division is valid only if cos x is not equal to 0!

$\sin x = 1$
And if sin x= 1 (or -1), then cos x= 0! Your whole method of solution is invalid.

Now, when $\sin x = 1$ then x is $\pi / 2$. But the answer is also $3 \pi / 2$

But isn't $3 \pi / 2 = -1$?

I'm missing a part of the theory obviously, was just wondering if someone could clarify.

Thank you!
Far better to use mr. fantastic's method, which does not depend on dividing by a function.

6. Thank you everyone for the help. I see what I was doing wrong now!