PrecálculoIdentities and EquationsF-TF.B.7

Solving Basic Trig Equations

Find every solution in one period, then all solutions.

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Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.

An identity is true for every angle; an equation like $2\sin x - 1 = 0$ is true only for some. Solving one uses everything from reference angles and inverse trig functions. The new part is finding all the solutions.

Idea clave

Isolate the trig function. Find the reference angle from the size of the value. List every angle in one period, $[0, 2\pi)$, in the quadrants where the sign is right. For all solutions, add whole periods: $+2k\pi$ for sine and cosine, $+k\pi$ for tangent.

Por qué funciona

On the unit circle, $\sin x = \frac{1}{2}$ means the point is at height $\frac{1}{2}$. The horizontal line $y = \frac{1}{2}$ crosses the circle at two points, one in Quadrant I and its mirror image in Quadrant II. A calculator's $\sin^{-1}$ can only report one of them, the one in its range. Then every full turn repeats both.
The line $y = \frac{1}{2}$ crosses the sine wave twice in every period of $2\pi$.

Worked examples

Ejemplo resuelto: Two solutions

Solve $2\sin x - 1 = 0$ on $[0, 2\pi)$, then give all solutions.
  1. Isolate: $\sin x = \frac{1}{2}$.
  2. Reference angle $\frac{\pi}{6}$. Sine is positive in Quadrants I and II: $x = \frac{\pi}{6}$ and $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
  3. All solutions: $x = \frac{\pi}{6} + 2k\pi$ or $x = \frac{5\pi}{6} + 2k\pi$, for any integer $k$.

Respuesta: $\frac{\pi}{6}, \frac{5\pi}{6}$ in $[0, 2\pi)$

Ejemplo resuelto: A negative value

Solve $2\cos x + \sqrt{3} = 0$ on $[0, 2\pi)$.
  1. $\cos x = -\frac{\sqrt{3}}{2}$. Reference angle $\frac{\pi}{6}$.
  2. Cosine is negative in Quadrants II and III: $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$ and $\pi + \frac{\pi}{6} = \frac{7\pi}{6}$.

Respuesta: $\frac{5\pi}{6}, \frac{7\pi}{6}$

Ejemplo resuelto: Tangent

Solve $\tan x = -1$ on $[0, 2\pi)$.
  1. Reference angle $\frac{\pi}{4}$; tangent is negative in Quadrants II and IV.
  2. $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$, exactly $\pi$ apart, as tangent's period predicts.

Respuesta: $\frac{3\pi}{4}, \frac{7\pi}{4}$

Error común

Stopping at the one answer a calculator gives. $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$ is only one solution. Draw the horizontal (for sine) or vertical (for cosine) line on the unit circle and count where it crosses: usually twice per turn. Check each answer by substituting it back.
Same height, other side of the circle.

Práctica

Apruebas con 3 respuestas correctas, las 2 últimas seguidas. Cada respuesta correcta da 2 XP.

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Fuentes y lecturas

  • OpenStax Precalculus 2e, Section 7.5 Solving Trigonometric Equations
  • Common Core State Standards for Mathematics, F-TF.B.7