What does it mean to solve a trigonometric equation over the interval \([0, 2\pi)\)?
It means finding all angle values in radians within one full counterclockwise revolution of the unit circle, starting at \(0\) and strictly less than \(2\pi\), that satisfy the equation.
Isolate the function:
$$ \sin(x) = \frac{1}{2} $$
Sine is positive in Quadrants I and II. The reference angle is \(\frac{\pi}{6}\).
Therefore, \(x = \frac{\pi}{6}\) or \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\).
Isolate the function:
$$ \cos(x) = -\frac{\sqrt{3}}{2} $$
Cosine is negative in Quadrants II and III with reference angle \(\frac{\pi}{6}\).
Therefore, \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\) or \(x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\).
Isolate the function:
$$ \tan(x) = -1 $$
Tangent is negative in Quadrants II and IV with reference angle \(\frac{\pi}{4}\).
Therefore, \(x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\) or \(x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}\).
Solve for \(x \in [0, 2\pi)\):
$$ \cos(x) = 0 $$
Cosine corresponds to the \(x\)-coordinate on the unit circle. The \(x\)-coordinate is zero on the vertical axis at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\).
Solve for \(x \in [0, 2\pi)\):
$$ \sin(x) = -1 $$
Sine corresponds to the \(y\)-coordinate on the unit circle. The \(y\)-coordinate equals \(-1\) only at the bottom of the unit circle, which occurs at \(x = \frac{3\pi}{2}\).
Solve for \(x \in [0, 2\pi)\):
$$ \sec(x) = -2 $$
Convert to reciprocal function:
$$ \cos(x) = -\frac{1}{2} $$
Cosine is negative in Quadrants II and III with reference angle \(\frac{\pi}{3}\).
Therefore, \(x = \frac{2\pi}{3}\) or \(x = \frac{4\pi}{3}\).
Convert to reciprocal function:
$$ \sin(x) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$
Sine is positive in Quadrants I and II with reference angle \(\frac{\pi}{4}\).
Therefore, \(x = \frac{\pi}{4}\) or \(x = \frac{3\pi}{4}\).
Why does the equation \(\sin(x) = 2.5\) have no real solutions?
The range of the basic sine function is \([-1, 1]\). Because \(2.5 > 1\), there is no real angle \(x\) whose sine equals \(2.5\).
What is a reference angle, and how is it used to solve trigonometric equations?
The reference angle \(\theta_{\text{ref}}\) is the acute positive angle between the terminal ray and the \(x\)-axis. Once \(\theta_{\text{ref}}\) is found, the signs of the trig function determine which quadrants contain the valid terminal sides.
