How would you solve a trigonometric equation (using the unit circle), which includes a negative domain, such as:
$$\sin(x) = 1/2, \text{ for } -4\pi < x < 4\pi$$
I understand, that the sine function is positive in the 1st and 2nd quadrants of the unit circle, so to calculate the solutions in the positive domain it's:
$\theta$ (1st quadrant), $\pi – \theta$ (2nd quadrant), $2\pi + \theta$ (returning to the 1st quadrant), $3\pi – \theta$ (returning the the 2nd quadrant).
Though, how do you traverse the unit circle for the negative domain, is there a pattern like for the positive solutions $(\theta, \pi-\theta, \pi+\theta, 2\pi-\theta$, and so on)
Best Answer
$$\sin x=\frac12\iff x=\begin{cases}\frac\pi3+2k\pi\\{}\\\frac{2\pi}3+2k\pi\end{cases}\;,\;\;\;\;\;k\in\Bbb Z$$
And now just solve the inequalities on $\;k\;$:
$$\begin{cases}-4\pi<\frac\pi3+2k\pi<4\pi\iff -\frac{13}3<2k<\frac{11}3\iff -\frac{13}6<k<\frac{11}6\\{}\\-4\pi<\frac{2\pi}3+2k\pi<4\pi\iff -\frac{14}3<2k<\frac{10}3\iff -\frac73<k<\frac53\end{cases}\iff$$
$$k=-2,-1,0,1,$$