Question

Help?

Find all solutions in the interval [0, 2π).
sin2x - cos2x = 0

Answer 1

sin2x = cos2x
sin2x = sin(pi/2-2x)
is last step clear to you?

Answer 2

Oh, crud. I'm sorry. The equation is:
\[\sin^2 x = \cos^2 x\]

Answer 3

you could do a few things: use sin^2 = 1 - cos^2 and put everything in terms of cos (or sin)
or divide by cos^2 and solve \(\tan^2 x =1\)

Answer 4

@This_Is_Batman
\[\sin^2x-\cos^2x = 0\]
\[\cos^2x-\sin^2x=0\]
we know that \[\cos^2x-\sin^2x = \cos2x\]
therefore, \[\cos2x = 0 \]
is all steps clear to you?

Answer 5

Uhmm... I think so..

Answer 6

How do I find the solutions????

Answer 7

\[\cos2x = \cos(\frac{ n \pi }{ 2 })\]
where n is odd number
\[2x = \frac{ n \pi }{ 2}\]
\[x=\frac{ n \pi }{4 }\]

Answer 8

What?!

Answer 9

do u know how to make cosine graph?

Answer 10

graph is not drawn very continuously apologies for that
from the graph it can be concluded that
cosine is zero at pi/2 , 3pi/2, 5pi/2 and so on
so i write it down as a general formula cos(npi/2)
where n= 1,3,5,7,so on