Solving trigonometric equations

Why there are always many solutions

• Trigonometric functions repeat, so an equation such as $\sin\theta = 0.5$ has infinitely many solutions.
• The calculator gives you only one — the principal value from the inverse function. Every other solution must be found by reasoning.
• The principal value ranges:
  • $\sin^{-1}$ returns an angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
  • $\cos^{-1}$ returns an angle in $\left[0, \pi\right]$
  • $\tan^{-1}$ returns an angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Where the other solutions come from

• The CAST rule records where each function is positive, reading anticlockwise from the fourth quadrant:
  • C — 4th quadrant: cos positive
  • A — 1st quadrant: all positive
  • S — 2nd quadrant: sin positive
  • T — 3rd quadrant: tan positive

• Within one revolution $[0, 2\pi)$, each equation has two solutions (unless the value is exactly at a maximum, minimum or zero):

where $\alpha$ is the principal value (adjusted into the interval if negative).

• Tangent is the exception: because its period is $\pi$, its solutions are $\pi$ apart, not paired symmetrically.

The method

• Isolate the trigonometric function so it stands alone.

• Find the principal value with the inverse function.

• Use the CAST rule or a sketch to find the other solution in one revolution.

• Add or subtract multiples of the period to fill the required interval.

• Check every solution lies inside the stated interval, and that none is missing.

• Worked through — solve $2\sin\theta = 1$ for $0 \leq \theta \leq 2\pi$:

  • Isolate: $\sin\theta = 0.5$
  • Principal value: $\theta = \sin^{-1}(0.5) = \frac{\pi}{6}$
  • Sine is positive in quadrants 1 and 2, so the second solution is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$
  • Solutions: $\theta = \frac{\pi}{6}, \; \frac{5\pi}{6}$

When the value is negative

• A negative value shifts which quadrants apply.
• For $\cos\theta = -0.5$ on $[0, 2\pi)$:
  • $\cos^{-1}(-0.5) = \frac{2\pi}{3}$ — the calculator already returns a second-quadrant angle
  • Cosine is negative in quadrants 2 and 3, so the other solution is $2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}$
• Sketching the curve and drawing the horizontal line $y = k$ shows immediately how many intersections there are and roughly where.

Equations with a multiple angle

• For $\sin 3\theta = 0.5$ on $0 \leq \theta \leq 2\pi$, the argument $3\theta$ runs from $0$ to $6\pi$ — three full revolutions.
• The method:
  • Widen the interval for the argument: if $0 \leq \theta \leq 2\pi$, then $0 \leq 3\theta \leq 6\pi$.
  • Find every solution for $3\theta$ in that widened interval.
  • Divide each by 3 at the end.
• This is where solutions get lost. A multiple angle multiplies the number of solutions, so $\sin 3\theta = 0.5$ has six solutions in one revolution of $\theta$, not two.

Quadratic trigonometric equations

• An equation such as $2\sin^2\theta - \sin\theta - 1 = 0$ is a quadratic in $\sin\theta$.
• Substitute $u = \sin\theta$, factorise, then solve each resulting equation separately.
• $(2u + 1)(u - 1) = 0$ gives $\sin\theta = -\frac{1}{2}$ or $\sin\theta = 1$, and each is solved as usual.
• Reject any value outside $[-1, 1]$ for sine or cosine — it has no solution.

Using an identity first

• If an equation contains two different functions, use an identity to reduce it to one.
• $2\cos^2\theta + \sin\theta = 2$ becomes, using $\cos^2\theta = 1 - \sin^2\theta$:
  • $2 - 2\sin^2\theta + \sin\theta = 2$, so $\sin\theta(1 - 2\sin\theta) = 0$
• Factorise rather than dividing, so no solution is lost.