When cos(x) + sin(x) = 0 I can see how cos(x) cannot equal 0.

But when written as cos(x)[1 + tan(x)] = 0 cos(x) can be 0?

Could someone please try to explain this to me :)

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- Jan 9th 2013, 07:21 AMkinhew93General trig question
When cos(x) + sin(x) = 0 I can see how cos(x) cannot equal 0.

But when written as cos(x)[1 + tan(x)] = 0 cos(x) can be 0?

Could someone please try to explain this to me :) - Jan 9th 2013, 07:42 AMebainesRe: General trig question
- Jan 9th 2013, 08:20 AMSorobanRe: General trig question
Hello, kinhew93!

Quote:

$\displaystyle \text{When }\,\cos x + \sin x \:=\:0,\,\text{ I can see that }\cos x \ne 0$

$\displaystyle \text{But when written as: }\,\cos x(1 + \tan x) \:=\:0,\: \cos x\text{ can be 0.}$

$\displaystyle \text{Could someone please try to explain this to me?}$

We have: .$\displaystyle \sin x \:=\:-\cos x$

. . . . . . . . $\displaystyle \frac{\sin x}{\cos x} \:=\:-1$ . . . You are right: $\displaystyle \cos x \ne 0$

. . . . . . . . $\displaystyle \tan x \:=\:-1$

Therefore: n . $\displaystyle x \:=\:-\tfrac{\pi}{4} + \pi n$