Multiple Angle Trigonometric Equations

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Techniques for solving equations containing arguments of the form kx over bounded intervals.

Multiple Angle Trigonometric Equations

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When solving $\sin(2x) = \frac{1}{2}$ for $x \in [0, 2\pi)$, what happens to the domain of the argument $2x$?

If $0 \le x < 2\pi$, then multiplying by $2$ expands the argument range to $0 \le 2x < 4\pi$. You must find all solutions across two full cycles before dividing by $2$.

Solve for $x \in [0, 2\pi)$: $$ \sin(2x) = \frac{\sqrt{3}}{2} $$

For $2x \in [0, 4\pi)$: $$ 2x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3} $$ Divide each value by $2$: $$ x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3} $$

Solve for $x \in [0, 2\pi)$: $$ \cos(3x) = 0 $$

For $3x \in [0, 6\pi)$, cosine is zero at odd multiples of $\frac{\pi}{2}$: $$ 3x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2} $$ Divide by $3$: $$ x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} $$

Solve for $x \in [0, 2\pi)$: $$ 2\cos(2x) + 1 = 0 $$

Isolate the function: $$ \cos(2x) = -\frac{1}{2} $$ In $[0, 4\pi)$, reference angle is $\frac{\pi}{3}$ in Quadrants II and III: $$ 2x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3} $$ Divide by $2$: $$ x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} $$

Solve for $x \in [0, 2\pi)$: $$ \tan(2x) = 1 $$

For $2x \in [0, 4\pi)$, $\tan(2x) = 1$ when the reference angle is $\frac{\pi}{4}$ in Quadrants I and III: $$ 2x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4} $$ Divide by $2$: $$ x = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8} $$

Solve for $x \in [0, 2\pi)$: $$ \sin\left(\frac{x}{2}\right) = \frac{\sqrt{2}}{2} $$

For $x \in [0, 2\pi)$, the argument is $\frac{x}{2} \in [0, \pi)$. Within $[0, \pi)$, sine equals $\frac{\sqrt{2}}{2}$ at: $$ \frac{x}{2} = \frac{\pi}{4}, \frac{3\pi}{4} $$ Multiply by $2$: $$ x = \frac{\pi}{2}, \frac{3\pi}{2} $$

Solve for $x \in [0, 2\pi)$: $$ \cos\left(2x - \frac{\pi}{4}\right) = 1 $$

Cosine equals $1$ at integer multiples of $2\pi$: $$ 2x - \frac{\pi}{4} = 0, 2\pi, 4\pi $$ Add $\frac{\pi}{4}$: $$ 2x = \frac{\pi}{4}, \frac{9\pi}{4} $$ (The next value $\frac{17\pi}{4}$ yields $x > 2\pi$). Divide by $2$: $$ x = \frac{\pi}{8}, \frac{9\pi}{8} $$

How many total solutions in $[0, 2\pi)$ are expected for $\sin(B x) = c$ when $|c| < 1$ and $B$ is a positive integer?

Graph of sine of 2x over the interval 0 to 2 pi intersecting a horizontal line at four points, illustrating that two full cycles produce four solutions.

Each standard period $2\pi$ of $\sin(x)$ has $2$ solutions. The function $\sin(Bx)$ completes $B$ full periods in $[0, 2\pi)$, so there are typically $2B$ solutions.

Multiple Angle Trigonometric Equations

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