Why do trigonometric equations generally have infinitely many real solutions if unconstrained?
Trigonometric functions are periodic: their values repeat identically every cycle of length \(2\pi\) (or \(\pi\) for tangent and cotangent). Adding any integer multiple of the period generates another valid solution.
What does the integer parameter \(k\) (or \(n\)) represent in general solution formulas?
The parameter \(k \in \mathbb{Z}\) represents any integer (positive, negative, or zero), indicating the number of full periods added to or subtracted from a base solution.
Find the general solution in radians for:
$$ \cos(x) = \frac{1}{2} $$
Base solutions in \([0, 2\pi)\) are \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\) (or \(\pm\frac{\pi}{3}\)).
Adding integer multiples of the period \(2\pi\):
$$ x = \pm\frac{\pi}{3} + 2\pi k, \quad k \in \mathbb{Z} $$
Find the general solution in radians for:
$$ \sin(x) = \frac{\sqrt{3}}{2} $$
Base solutions are \(x = \frac{\pi}{3}\) and \(x = \frac{2\pi}{3}\).
Adding multiples of the period \(2\pi\):
$$ x = \frac{\pi}{3} + 2\pi k \quad \text{or} \quad x = \frac{2\pi}{3} + 2\pi k, \quad k \in \mathbb{Z} $$
Why can the general solution to \(\tan(x) = c\) be written with a single term using \(+\pi k\) instead of two terms with \(+2\pi k\)?
The tangent function has a period of \(\pi\) rather than \(2\pi\). Solutions in opposite quadrants are exactly \(\pi\) radians apart, so adding \(\pi k\) captures all solutions in one statement.
Find the general solution in radians for:
$$ \tan(x) = \sqrt{3} $$
One base angle is \(x = \frac{\pi}{3}\). Since tangent repeats every \(\pi\) radians:
$$ x = \frac{\pi}{3} + \pi k, \quad k \in \mathbb{Z} $$
Find the general solution in radians for:
$$ \sin(x) = 0 $$
On the unit circle, sine is zero at \(0\), \(\pi\), \(2\pi\), \(3\pi\), and so on.
Compact general formula:
$$ x = \pi k, \quad k \in \mathbb{Z} $$
Find the general solution in radians for:
$$ \cos(x) = 0 $$
Cosine is zero at odd multiples of \(\frac{\pi}{2}\).
Compact general formula:
$$ x = \frac{\pi}{2} + \pi k, \quad k \in \mathbb{Z} $$
Express the general solution of \(\cos(x) = -1\) using integer \(k\).
Cosine equals \(-1\) at \(\pi\), \(3\pi\), \(5\pi\), etc.
General solution:
$$ x = \pi + 2\pi k, \quad k \in \mathbb{Z} $$
