d the exact Solutions of the ien equation, in radians With \( 0 \leq x<2 \pi \). \( -2 \sin x=1 \)

To solve the equation \(-2 \sin x = 1\), we first isolate \(\sin x\) by dividing both sides by -2, yielding \(\sin x = -\frac{1}{2}\). The sine function is negative in the third and fourth quadrants, so we look for angles where sine achieves that value. The reference angle corresponding to \(\frac{1}{2}\) is \(\frac{\pi}{6}\). Therefore, the solutions within the interval \(0 \leq x < 2\pi\) are: - In the third quadrant: \(x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\) - In the fourth quadrant: \(x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}\) So, the exact solutions are \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\). In trigonometry, understanding the unit circle can really help you visualize these solutions. Remember that the angles corresponding to \( \sin x = -\frac{1}{2} \) are found by looking where the sine function dips below the x-axis on the circle. Each quadrant has its own unique signs for sine, so be sure to reference that when solving similar equations! Experimenting with graphing the sine function can be a fantastic way to build intuition about its behavior! By plotting \(y = \sin x\) and horizontally drawing a line at \(y = -\frac{1}{2}\), you can visually identify the intersection points, which correspond to the angles where the sine value meets the equation. Keep in mind that mistakes often arise from overlooking the signs; double-checking quadrants is an essential step in solving these types of equations!