# Principal solution of $\sin(x)=1$ from $-2 \pi$ to $2 \pi$

Solve $\sin(x)=1$ for values of $x$ where $-2\pi\le x\le 2\pi$

Now, I know that $sin(\pi/2$)=$1$ in 1st quadrant and by using $sin(\pi-x)=sin(x)$ I still have $\pi/2$ and by using $sin(2\pi-x)=sin(x)$ I have $3\pi/2$.But the answer says the principal solutions are $-3\pi/2$ and $\pi/2$

I am very weak in trigonometry so I don't seem to understand the values for $0$ to $-2\pi$ and an explanation using basic identities would be helpful

(although this might be irrelevant but I came across this doubt while solving a question for Tangents and Normals under Application Of Derivatives)

http://www.ncert.nic.in/ncerts/l/lemh106.pdf Page237, or page 44 of the PDF

• $sin(2\pi-x)=-sin(x)$ Feb 12, 2016 at 16:36

$\sin(2\pi - x) = \sin x$ isn't a true identity; I think you're thinking of $\sin(x + 2\pi) = \sin x$. This identity also gives $\sin(x - 2\pi) = \sin x$, which gives you the solution $x = -3\pi / 2$ from $x = \pi / 2$.
To understand the interval $[-2\pi, 2\pi]$, it's essentially two loops around the unit circle. $[0, 2\pi]$ is the classical unit circle, starting at $\theta = 0$ (the $+x$ direction on the $xy$ plane) and moving counterclockwise, completing the circle at $\theta = 2\pi$. $[-2\pi, 0]$ is traversing the circle clockwise.
• But what about $x=3\pi/2$ and $x=-\pi/2$ Feb 12, 2016 at 16:38
• $\sin (3\pi / 2) = \sin (-\pi/2) = -1$. Feb 12, 2016 at 16:39
• Could you clarify a little bit more for the range $-2\pi$ and $2\pi$ because I think I have forgotten and jumbled up the identities. Feb 12, 2016 at 16:45
• That does it...just one more thing,when I have $sin(x)=1$ do I need to test the values for both the first and second quadrant,as $sin(x)$ is positive in 1st and 2nd both but the answer evaluates to $x=\pi/2$ in 1st and same in 2nd quadrant by using $sin(\pi-x)$ or is it the same thing? Feb 12, 2016 at 16:54