Question

How do you solve for sinx=cosx [0,2pi] ?

Answer 1

\[\sin x=\cos x\\
\frac{\sin x}{\cos x}=\frac{\cos x}{\cos x}\\
\tan x=1\]
When is tangent equal to 1?

Answer 2

that would be at pi/4 and 5pi/4...right?

Answer 3

which is kind of like asking "where is sine equal to cosine"

Answer 4

Yes

Answer 5

OH!!!

Answer 6

funny how restating a problem makes it easier right?

Answer 7

Just remember... it isn't always "safe" to divide by something like cos(x). But you can get away with it here, since the solution set requires that sin(x)=cos(x) and that never happens anywhere that cos(x)=0. :)

Answer 8

yeah LOL

oh, and how do I find the period of the following trig equations?
y=-3tan pi x

-3sec(- 6x) +2

Answer 9

Just don't try it if you have:

\(\Large \cos^2(x)=cos(x)\) :)

Answer 10

try it?

Answer 11

dividing by cos(x)

Answer 12

try what? Is that impossibl?

Answer 13

*impossible to solve?

Answer 14

why is it bad to do that?

Answer 15

the period of
\[\tan(bx)\] is \(\frac{\pi}{b}\)

Answer 16

If B is the coefficient on x (e.g., if you have y=sin(2x) then B=2) then:

For sine, cosine, cosecant and secant, period = 2pi/B
For tangent and cotangent, period = pi/B

Answer 17

Sorry, @lucy4104 , did you read my first comment above also? I didn't mean to confuse you, was just saying that you can't ALWAYS go dividing an equation by an unknown (like cos(x)) but it was OK here. In a different equation, you can't do it!

Answer 18

why not...?

Answer 19

Because what if cos(x)=0? :)

Answer 20

\[\Large \cos^2(x)=\cos(x)\]\[\Large \cos^2(x)-\cos(x)=0\]\[\Large \cos(x)(\cos(x)-1)=0\]
So cos(x)=0, hence x=pi/2 or 3pi/2.... OR
cos(x)=1, hence x=0

That's 3 solutions, but if you start THIS one by dividing by cos(x), you end up losing the first 2 solutions, because you eliminate the cos(x)=0 part of the solution.

Answer 21

cos(x)=0... then x is pi/2 +n(pi)

Answer 22

But it was OK to do so for YOUR equation above, since when cos(x)=sin(x), cos(x) is never = 0.

Answer 23

Yes, if you want the general form solution set with ALL real numbers.

Answer 24

oh, ok then you're saying to watch out? Wait, then how can you tell if its ok?

Answer 25

In general, don't divide by an unknown if it COULD be =0. But in your original equation

cos(x)=sin(x)

the solution set can't include any values where cos(x)=0, since that is true (x=pi/2 and x=3pi/2 and all coterminals) ONLY at angles where cos(x) is NOT = sin(x). So you can divide by cos(x) because it can't =0.

Don't get too hung up on it. :) it was kind of an aside. :)

Answer 26

It's just like in algebra, if you solve

\(x^2=x\) you do it by setting = 0 and factoring, not by dividing by x. Same idea.

Answer 27

oh, ok now i c it~ That was a good way to put it. Thank you!