Question

solve -sin(x)+cos(x)=0

Answer 1

Add sin(x) to both sides

Answer 2

What's the range of x?

Answer 3

That just changes the format, the range of X is (0,2pi)

Answer 4

cos(x)=sin(x)

Answer 5

Change sin(x) to cos(pi/2-x)

Answer 6

Why would you do that?

Answer 7

Because sin(x)=cos(pi/2-x)

Answer 8

oh, so you would change sin(x) then solve it?

Answer 9

Yep

Answer 10

Let me try that quick

Answer 11

Mind showing us your steps?

Answer 12

well, now that I am thinking, I am not sure how you would solve it...

Answer 13

There are actually 3 methods to solve this

Answer 14

Method 1:
cosx-sinx=0
cosx=sinx
cosx=cos(pi/2-x)
x=pi/2-x OR x=2pi+pi/2-x
2x=pi/2 OR 2x=5pi/2
x=pi/4 OR x=5pi/4

Method 2:
cosx-sinx=0
cosx=sinx
cos^2x=sin^2x
cos^2x-sin^2x=0
cos(2x)=0
2x=pi/2 OR 2x=3pi/2 OR 2x=5pi/2
x=pi/4 OR x=3pi/4(rej) OR x=5pi/4

Method 3:
cosx-sinx=0
cos^2x-2sinxcosx+sin^2x=0
-2sinxcosx=-1
sin(2x)=1
2x=pi/2 OR 2x=5pi/2
x=pi/4 OR 2x=5pi/4

Answer 15

Method 4 would be a modification of method 1 using sine instead of cosine

Answer 16

Or sin(x) = cos(x)
tan(x) = 1
x = arctan(1)

Answer 17

For what my problem is, method one would be best

Answer 18

Thanks for your help guys, much appreciated

Answer 19

no problem