March 19th, 2018, 06:27 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27  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: 2,039 Thanks: 1063 
This is solved when one term equals 1 and the other equals 0 and nowhere else. Thus $x = \left(\dfrac 1 2 +2 k\right) \vee (2k) ,~k \in \mathbb{Z}$ 
March 19th, 2018, 09:45 AM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27 
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: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond 
$\displaystyle \sin \pi x + \cos \pi x =1$ $\displaystyle (\sin \pi x + \cos \pi x)^2 =1$ $\displaystyle \sin^2\pi x+2\sin\pi x\cos\pi x+\cos^2\pi x=1$ $\displaystyle 1+2\sin\pi x\cos\pi x=1$ $\displaystyle 2\sin\pi x\cos\pi x=0\implies\sin\pi x=0\vee\cos\pi x=0$ 

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elementary, equation, trig 
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