How can I find the solutions of tan(x)=1?

Dec 2018
29
2
USA
The original equation was sin(x)=cos(x). The solutions can lie in the other variant, tan(x)=1 [sin(x)/cos(x) = tan(x) by default).

I looked at the sin/cos periodicity table and noticed one solution for both functions of cos(x) and sin(x): (pi/4, square root of 2/2). How could I find the other solutions; in other words, is there an equation to note them?
 

Plato

MHF Helper
Aug 2006
22,472
8,642
The original equation was sin(x)=cos(x). The solutions can lie in the other variant, tan(x)=1 [sin(x)/cos(x) = tan(x) by default).
The answer is that the $\cos(x)=\sin(x)$ is true if $x=\frac{\pi}{4}+2k\pi,~k\in\mathbb{Z}$
OR $\cos(x)=\sin(x)$ is true if $x=\frac{5\pi}{4}+2k\pi,~k\in\mathbb{Z}$
 
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Debsta

MHF Helper
Oct 2009
1,311
597
Brisbane
The answer is that the $\cos(x)=\sin(x)$ is true if $x=\frac{\pi}{4}+2k\pi,~k\in\mathbb{Z}$
OR $\cos(x)=\sin(x)$ is true if $x=\frac{5\pi}{4}+2k\pi,~k\in\mathbb{Z}$
There are solutions in the 1st and 3rd quadrants (where tan x is positive)
 
Dec 2018
29
2
USA
Thank you Plato, but how did you get the equations x= x/4 + 2kpi and x = 5pi/4 + 2kpi?
 

Plato

MHF Helper
Aug 2006
22,472
8,642
Thank you Plato, but how did you get the equations x= x/4 + 2kpi and x = 5pi/4 + 2kpi?
Well want $\tan(x)=\dfrac{\sin(x)}{\cos(x)}=1$.
That is a fundamental fact that you are expected to know.
For all $x$ it is true that $\sin(x)=\sin(x+2k\pi),~k\in\mathbb{Z}$ That is also the case for $\cos(x)$.
You are also expected to know that $\sin\left(\dfrac{\pi}{4}\right)=\cos\left(\dfrac{\pi}{4}\right)~\&~\sin\left(\dfrac{-3\pi}{4}\right)=\cos\left(\dfrac{-3\pi}{4}\right)$.
 
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May 2019
7
7
Kansas
Let x= sin(t) and y= cos(t). Then \(\displaystyle x^2+ y^2= sin^2(t)+ cos^2(t)= 1\) which has the unit circle as graph. You are also given that sin(t)= cos(t) so x= y. The graph of that is a straight line. (In fact, it is a diameter of the circle.) Both are satisfied where the straight line y= x crosses the circle \(\displaystyle x^2+ y^2= 1\). That is, \(\displaystyle x^2+ x^2= 2x^2= 1\) so \(\displaystyle x= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}\). One solution is, as you give, \(\displaystyle t= \frac{\pi}{4}\). The other is half way around the circle, \(\displaystyle \frac{\pi}{4}+ \pi= \frac{5\pi}{4}\)
 
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