Question

sin^2x-cos^2x=0; [0,2pi)

Answer 1

\[\sin ^2x-\cos ^2x=(\sin x-\cos x)(\sin x+\cos x)\]

Answer 2

i dont understand…

Answer 3

You don't know how to factor the difference of two squares or what?

Answer 4

no i got that part, but i don't really understand whats next

Answer 5

Set each factor equal to 0 and solve the resulting equation.

Answer 6

so…
sin^4x-2sin^2xcos^2x+cos^4x=0
then what

Answer 7

\[\sin x-\cos x=0\]
\[\sin x=\cos x\]
\[\frac{\sin x}{\cos x}=\frac{\cos x}{\cos x}=1\]
\[\tan x=1\]

Answer 8

what's next after my above comment?

Answer 9

I'm trying to figure out where you got that comment.

Answer 10

here's what i got…sin^4x-2sin^2cos^2+cos^4x=0
1-2sin^2xcos^2x=0
sin^2xcos^2x=1/2
sin^2x=1/2 and cos^2x=1/2
(pi/6, 5pi/6) and (pi/3,5pi/3)
am i doing it correctly so far?

Answer 11

I can't figure out where you got that expression sin^4x-2sin^2xcox^2x+cos^4x

Answer 12

i squared sin^2x-cos^2x like you said

Answer 13

Where did I say to square something?

Answer 14

oops, you said sin2x−cos2x=(sinx−cosx)(sinx+cosx)… so what do i do then?

Answer 15

Medals 0

Set each factor equal to 0 and solve the resulting equations.

Answer 16

what factors though… i don't understand where to get them

Answer 17

Do you know what factors are?

Answer 18

yes lol

Answer 19

What are the factors of x^2-9

Answer 20

3,-3

Answer 21

No. The factors are (x-3)(x+3)

Answer 22

If you have the equation x^2-9=0
First you factor

Answer 23

And then you set each factor equal to 0 and solve each equation

Answer 24

So factor sin^2x-cos^x

Answer 25

i just don't see how to factor that equation

Answer 26

idk if I'm having a brain fart or what but i don't know how to get them

Answer 27

k well thanks for the help…..

Answer 28

The same way you factored x^2-9

Answer 29

The same way I showed you earlier:

Answer 30

\[\sin ^2x-\cos ^2x=(\sin x-\cos x)(\sin x+\cos x)\]