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mathstudent55 Group TitleBest ResponseYou've already chosen the best response.0
dw:1359396748498:dw
 one year ago

kassandratb Group TitleBest ResponseYou've already chosen the best response.0
what do i do one i've done this
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
Well, remember that cos(2x) = cos^2 x  sin^2 x So if cos^2 x = sin^2 x then cos^2 x  sin^2 x = 0 or cos(2x) = 0 Can you take it from here?
 one year ago

kassandratb Group TitleBest ResponseYou've already chosen the best response.0
when I do this : cos^2xsin^2x=0 cos(2x)=0 cosx=0 x=pi, 2pi, +2k ?
 one year ago

kassandratb Group TitleBest ResponseYou've already chosen the best response.0
JamesJ i feel like i did this wrong
 one year ago

mathstudent55 Group TitleBest ResponseYou've already chosen the best response.0
dw:1359397774601:dw
 one year ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
You need to make a substitution. Replace sin^2 theta with 1  cos^2 theta or replace cos^2 theta with 1sin^2 theta.
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
If cos(2x) = 0 then 2x = pi/2 + k.pi, for any integer k. Hence x = pi/4 + k.pi/2, for any integer k.
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
You can see those solutions on this graph as well.
 one year ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
\[\sin^2\theta =1\sin ^2\theta \] \[2\sin ^2\theta =1\] \[\sin^2\theta =\frac{1}{2}\] \[\sin \theta = \pm \frac{\sqrt{2}}{2}\] \[\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} + 2\pi n\]
 one year ago

kassandratb Group TitleBest ResponseYou've already chosen the best response.0
Thank you all it's starting to make a lot more since to me now so I just gotta know my identities?
 one year ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
Yes. That is key.
 one year ago
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