anonymous
  • anonymous
Help solving cos^2 (2x) -sin^2 (2x)=0 and sin(2x)=sin x ?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
arindameducationusc
  • arindameducationusc
Do you have options? i guess I found the first answer... but need to check if it matches @Kiritina
anonymous
  • anonymous
cos^2 (2x) -sin^2 (2x) = cos(4x)
anonymous
  • anonymous
No answers available, sorry. I'm not sure where to start with either equation.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

arindameducationusc
  • arindameducationusc
can you manage now @Kiritina or should i do the solution?
anonymous
  • anonymous
I'd love it if you helped with the solution!
Jhannybean
  • Jhannybean
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?
anonymous
  • anonymous
https://gyazo.com/976ec9f78a10f5487c8421aaa06e36a0
arindameducationusc
  • arindameducationusc
O yes @Jhannybean is right I just solved the black portion I didn't think about the red!
arindameducationusc
  • arindameducationusc
ok both are different problems hehe :D
Jhannybean
  • Jhannybean
So you're looking for a general solution... x_x I'm terrible at finding those.
arindameducationusc
  • arindameducationusc
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
arindameducationusc
  • arindameducationusc
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
arindameducationusc
  • arindameducationusc
@Kiritina if you liked my response, dont forget to medal and fan. Thank you

Looking for something else?

Not the answer you are looking for? Search for more explanations.