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)
Page237, or page 44 of the PDF
$\endgroup$ 11 Answer
$\begingroup$$\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.
$\endgroup$ 6
