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anonymous
 4 years ago
Can you solve this for me?
sin((pi/6)+x)=cos((pi/4)x)
anonymous
 4 years ago
Can you solve this for me? sin((pi/6)+x)=cos((pi/4)x)

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myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2\[\sin(x)=\cos(\frac{\pi}{2}x) ; \cos(x)=\sin(\frac{\pi}{2}x)\] \[\sin(\frac{\pi}{6}+x)=\sin(\frac{\pi}{2}(\frac{\pi}{4}x))\] \[\cos(\frac{\pi}{4}x)=\cos(\frac{\pi}{2}(\frac{\pi}{6}+x))\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2i used those two top equations to rewrite what you have

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2no I suppose we should put something in there like 2npi

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2\[\sin(\frac{\pi}{6}+x)=\sin(\frac{\pi}{2}(\frac{\pi}{4}x)+2n \pi)\] \[\cos(\frac{\pi}{4}x)=\cos(\frac{\pi}{2}(\frac{\pi}{6}+x)+2n \pi)\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2where n is an integer

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2now to solve both equations set the insides equal and solve for x

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sin(\frac{\pi}{6}+x)=\cos(\frac{\pi}{4}x)\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sin\frac{\pi}{6}cosx+\cos\frac{\pi}{6}sinx=\cos\frac{\pi}{4}cosx+\sin\frac{\pi}{4}sinx\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx=\frac{\sqrt{2}}{2}cosx+\frac{\sqrt{2}}{2}sinx\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0\[(\frac{\sqrt{3}\sqrt{2}}{2})sinx=(\frac{\sqrt{2}1}{2})cosx\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0\[sinx=(\frac{\sqrt{2}1}{\sqrt{3}\sqrt{2}})cosx\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.0\[tanx=\frac{\sqrt{2}1}{\sqrt{3}\sqrt{2}}\]
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