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

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

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.1Well, 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?

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

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

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

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

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

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

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

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