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nphuongsun93
 2 years ago
Best ResponseYou've already chosen the best response.0solve for y, sorry

nphuongsun93
 2 years ago
Best ResponseYou've already chosen the best response.0gonna read this later, bbl 2 hours **

lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.0<will reply in 2 hours

RadEn
 2 years ago
Best ResponseYou've already chosen the best response.0i think not enough information for solve it

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0@nphuongsun93 What you tried? (just asking)

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0\(\cos x + \cos y  \cos x \cos y + \sin x \sin y=\frac{3}{2}\)

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0@asnaseer sir any ideas?

Coolsector
 2 years ago
Best ResponseYou've already chosen the best response.0i thought of maybe using cos(a) = [e^(ia) + e^(ia)] / 2 but i dont think it works either

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2Maybe try this:\[\cos(x)+\cos(y)=2\cos(\frac{x+y}{2})\cos(\frac{xy}{2})\]and:\[\cos(x+y)=2\cos^2(\frac{x+y}{2})1\]That will lead to:\[\cos(\frac{x+y}{2})\cos(\frac{xy}{2})\cos^2(\frac{x+y}{2})=\frac{1}{4}\]

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2which is a quadratic in \(\cos(\frac{x+y}{2})\)  although it also contains a \(\cos(\frac{xy}{2})\)

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2Solving the quadratic leads to:\[\cos(\frac{x+y}{2})=\frac{\cos(\frac{xy}{2})\pm\sqrt{\cos^2(\frac{xy}{2})1}}{2}\]

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2Which simplifies to:\[\cos(\frac{x+y}{2})=\frac{\cos(\frac{xy}{2})\pm\sin(\frac{xy}{2})}{2}\]

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2Don't know if this is helpful or not but I have to go now  good luck!

nphuongsun93
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks a lot @asnaseer i get it, that's just what i needed

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0@nphuongsun93 , what and how did you get the answer?

nphuongsun93
 2 years ago
Best ResponseYou've already chosen the best response.0\[\cos(\frac{x+y}{2})\cos(\frac{xy}{2})\cos^2(\frac{x+y}{2})0.25 = 0\] then let \[a = \cos(\frac{x+y}{2}) \space and \space b = \cos ( \frac{xy}{2})\] \[ab  a^2  0.25=0\] taking this as a quadratic in p, b^2 1 has to be positive to get real solution for a, but the highest value of b = cos((xy)/2) is 1 (angles), so the only real solution is b =1 or b^2 1 = 0 \[b= \cos(\frac{xy}{2})=1 \space or \space \frac{xy}{2}=0 \space or \space x=y\]
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