7.5 Solving Trigonometric Equations

Trigonometric equations ask you to find the angle(s) that make an equation involving trig functions true. Because trig functions are periodic, these equations often have infinitely many solutions, so you'll need to know how to find all of them within a given interval. This topic pulls together your algebra skills and your knowledge of trig identities from earlier in the unit.

Solving Basic Trigonometric Equations

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Solutions for basic trigonometric equations

The general strategy mirrors solving any equation: isolate the trig function, then "undo" it with an inverse.

1. Isolate the trig function on one side of the equation using addition, subtraction, multiplication, or division.

2. Apply the inverse trig function to find the angle.

   • For sine equations, use $\sin^{-1}$ (also written $\arcsin$).
   • For cosine equations, use $\cos^{-1}$ (also written $\arccos$).
   • For tangent equations, use $\tan^{-1}$ (also written $\arctan$).

3. Check that a solution exists. Sine and cosine both have a range of $[-1, 1]$, so an equation like $\sin\theta = 3$ has no solution.

4. Find all solutions in the given interval, typically $[0, 2\pi)$ or $[0°, 360°)$. Because trig functions repeat, there are usually two angles per period that satisfy the equation. Use the unit circle or reference angles to locate them.

Example: Solve $2\sin\theta - 1 = 0$ on $[0, 2\pi)$.

1. Isolate: $\sin\theta = \frac{1}{2}$
2. The reference angle is $\frac{\pi}{6}$ (since $\sin\frac{\pi}{6} = \frac{1}{2}$).
3. Sine is positive in Quadrants I and II, so $\theta = \frac{\pi}{6}$ and $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

Algebraic techniques in trigonometric equations

These are the same moves you'd make in any algebra problem, just applied to trig expressions.

• Add or subtract terms to get the trig function by itself on one side.
• Multiply or divide by a non-zero constant to remove coefficients.
• Once isolated, apply the inverse trig function to get the principal value (the single angle your calculator returns):
  • $\sin^{-1}$ returns values in $\left[-\frac{\pi}{2},\, \frac{\pi}{2}\right]$
  • $\cos^{-1}$ returns values in $[0,\, \pi]$
  • $\tan^{-1}$ returns values in $\left(-\frac{\pi}{2},\, \frac{\pi}{2}\right)$
• From the principal value, use reference angles and knowledge of which quadrants the function is positive/negative to find all solutions in the interval.
• For the general solution (all solutions, not just those in one period), add integer multiples of the period: $\theta + 2\pi n$ for sine and cosine, or $\theta + \pi n$ for tangent, where $n$ is any integer.

Calculator use for trigonometric solutions

• Check your mode first. If the problem uses radians, your calculator must be in radian mode; if degrees, use degree mode. A wrong mode setting is one of the most common errors on tests.
• Isolate the trig function, then press the appropriate inverse button ($\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$).
• Your calculator gives only the principal value. You still need to find additional solutions using reference angles and the symmetry of the unit circle.

Advanced Trigonometric Equation Solving Techniques

Quadratic-form trigonometric equations

Some trig equations look like quadratics in disguise. The key is to recognize the pattern and use substitution.

Example: $2\sin^2\theta - \sin\theta - 1 = 0$

1. Substitute $u = \sin\theta$. The equation becomes $2u^2 - u - 1 = 0$.

2. Solve the quadratic by factoring, completing the square, or the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

   • Factoring: $(2u + 1)(u - 1) = 0$, so $u = -\frac{1}{2}$ or $u = 1$.

3. Substitute back: $\sin\theta = -\frac{1}{2}$ or $\sin\theta = 1$.

4. Solve each equation for $\theta$ in the given interval using inverse trig and reference angles.

5. Reject any impossible values. If the quadratic gives $u = 2$, that's outside $[-1, 1]$ for sine or cosine, so discard it.

This same approach works for equations in $\cos\theta$ or $\tan\theta$.

Identities in trigonometric equation solving

When an equation contains more than one trig function, identities let you rewrite everything in terms of a single function so you can solve it.

Identities you'll use most often:

• Pythagorean identity: $\sin^2\theta + \cos^2\theta = 1$ (and its rearrangements: $\sin^2\theta = 1 - \cos^2\theta$, etc.)
• Double angle formulas:
  • $\sin(2\theta) = 2\sin\theta\cos\theta$
  • $\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$
• Power-reducing (half-angle) formulas:
  • $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$
  • $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$

Strategy: Look at what functions appear in the equation. If you see both $\sin\theta$ and $\cos\theta$, try using the Pythagorean identity to convert to one function. If you see $\sin(2\theta)$ alongside $\sin\theta$, expand the double angle formula so everything is in terms of $\theta$. After simplifying, solve using the basic or quadratic techniques above.

Multiple angle trigonometric equations

Equations like $\sin(3\theta) = \frac{\sqrt{3}}{2}$ or $\cos(2\theta) = -1$ involve a multiple of the angle rather than $\theta$ alone.

1. Treat the entire argument as a single variable. Let $\phi = 3\theta$ (or $2\theta$, etc.).
2. Solve for $\phi$ in the interval that accounts for the multiplier. If you need $\theta \in [0, 2\pi)$ and the multiplier is 3, then $\phi \in [0, 6\pi)$.
3. Find all solutions for $\phi$ in that expanded interval by listing the base solutions and adding full periods ($2\pi$ for sine/cosine).
4. Divide each $\phi$ value by the multiplier to get $\theta$.

This step of expanding the interval is where most mistakes happen. If you forget to widen the interval, you'll miss solutions.

For equations that mix different angles (like $\sin\theta + \cos(2\theta) = 0$), use sum/difference or double angle formulas to rewrite everything in terms of a single angle first, then solve.

Applications of Right Triangle Trigonometry

Real-world applications of right triangle trigonometry

Word problems in this section follow a consistent process:

1. Read the problem and identify what you know (side lengths, angles) and what you need to find.

2. Draw a right triangle and label the known and unknown quantities.

3. Choose the right trig ratio based on which sides are involved relative to the angle:

   • $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$
   • $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
   • $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$

4. Write and solve the equation. Plug in known values, then isolate the unknown using algebra or an inverse trig function.

5. Interpret your answer. Check that the units make sense and the value is reasonable (e.g., a building height shouldn't come out negative).

Common setups include angles of elevation (looking up from horizontal) and angles of depression (looking down from horizontal). In both cases, the angle is measured from the horizontal line, not from the vertical.

Additional Trigonometric Concepts

Reciprocal trigonometric functions and graphing

The three reciprocal trig functions are defined by flipping the original ratios:

• Cosecant: $\csc\theta = \frac{1}{\sin\theta}$
• Secant: $\sec\theta = \frac{1}{\cos\theta}$
• Cotangent: $\cot\theta = \frac{1}{\tan\theta}$

Because you're dividing by the original function, reciprocal functions are undefined wherever the original function equals zero. For example, $\csc\theta$ is undefined at $\theta = 0, \pi, 2\pi, \ldots$ (where $\sin\theta = 0$), and those values produce vertical asymptotes on the graph.

When graphing reciprocal functions, pay attention to:

• Vertical asymptotes at every zero of the original function
• Period: $\csc\theta$ and $\sec\theta$ share the $2\pi$ period of sine and cosine; $\cot\theta$ has period $\pi$, like tangent
• No amplitude in the traditional sense, since these functions extend to $\pm\infty$
• Domain and range: The range of $\csc\theta$ and $\sec\theta$ is $(-\infty, -1] \cup [1, \infty)$

Graphing trig equations can also serve as a visual check on your algebraic solutions. The $x$-values where two trig graphs intersect correspond to the solutions of the equation.