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March 19th, 2018, 06:27 AM   #1
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Elementary trig Equation

$\displaystyle \sin \pi x + \cos \pi x =1$
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March 19th, 2018, 08:27 AM   #2
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This is solved when one term equals 1 and the other equals 0 and nowhere else.

Thus $x = \left(\dfrac 1 2 +2 k\right) \vee (2k) ,~k \in \mathbb{Z}$
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March 19th, 2018, 09:45 AM   #3
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I'm not being able to remember how I used to solve this before.

Last edited by skipjack; March 19th, 2018 at 12:02 PM.
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March 19th, 2018, 11:39 AM   #4
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$\displaystyle \sin \pi x + \cos \pi x =1$

$\displaystyle (\sin \pi x + \cos \pi x)^2 =1$

$\displaystyle \sin^2\pi x+2\sin\pi x\cos\pi x+\cos^2\pi x=1$

$\displaystyle 1+2\sin\pi x\cos\pi x=1$

$\displaystyle 2\sin\pi x\cos\pi x=0\implies\sin\pi x=0\vee\cos\pi x=0$
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