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 March 19th, 2018, 06:27 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 224 Thanks: 26 Elementary trig Equation $\displaystyle \sin \pi x + \cos \pi x =1$
 March 19th, 2018, 08:27 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,941 Thanks: 1008 This is solved when one term equals 1 and the other equals 0 and nowhere else. Thus $x = \left(\dfrac 1 2 +2 k\right) \vee (2k) ,~k \in \mathbb{Z}$ Thanks from idontknow
 March 19th, 2018, 09:45 AM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 224 Thanks: 26 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.
 March 19th, 2018, 11:39 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,807 Thanks: 1045 Math Focus: Elementary mathematics and beyond $\displaystyle \sin \pi x + \cos \pi x =1$ $\displaystyle (\sin \pi x + \cos \pi x)^2 =1$ $\displaystyle \sin^2\pi x+2\sin\pi x\cos\pi x+\cos^2\pi x=1$ $\displaystyle 1+2\sin\pi x\cos\pi x=1$ $\displaystyle 2\sin\pi x\cos\pi x=0\implies\sin\pi x=0\vee\cos\pi x=0$ Thanks from Country Boy and romsek

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