Question

Help solving cos^2 (2x) -sin^2 (2x)=0 and sin(2x)=sin x ?

Answer 1

Do you have options? i guess I found the first answer... but need to check if it matches @Kiritina

Answer 2

cos^2 (2x) -sin^2 (2x) = cos(4x)

Answer 3

No answers available, sorry. I'm not sure where to start with either equation.

Answer 4

can you manage now @Kiritina
or should i do the solution?

Answer 5

I'd love it if you helped with the solution!

Answer 6

lets see... \[\cos^2(2x) -\sin^2(2x) = 0\]\[\color{red}{\sin(2x) = \sin(x)}\] is the red portion what you're given? Do you mind posting the actual problem by a picture?

Answer 7

https://gyazo.com/976ec9f78a10f5487c8421aaa06e36a0

Answer 8

O yes @Jhannybean is right I just solved the black portion
I didn't think about the red!

Answer 9

ok both are different problems hehe :D

Answer 10

So you're looking for a general solution... x_x I'm terrible at finding those.

Answer 11

okay 1st question
we have cos^2x-sin^2x=0
=>cos4x=0
\[=>4x=\cos^{-1} (0)\]
\[=\cos^{-1} (\cos (\frac{ \pi }{ 2}))\]
=>4x=pi/2
>x=pi/8

Answer 12

2nd question,
sin2x=2sinxcosx
so,
sin2x=sinx
=>2sinxcox/sinx=1
=>2cosx=1
=>cosx=1/2
x=cos inverse(1/2)
x=cos inverse (cos pi/3)

x=pi/3

Answer 13

@Kiritina
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