Solving Trigonometric Equations
Basic Trigonometric Equations
Trigonometric equations involve trigonometric functions and are solved by finding the angles that satisfy the given condition.
Linear Trigonometric Equations
The simplest form of trigonometric equations are linear, such as:
$2 \sin x = 1, 0 \leq x \leq 2\pi$
To solve this:
- Isolate the trigonometric function: $\sin x = \frac{1}{2}$
- Find the reference angle: $x = \arcsin(\frac{1}{2}) = \frac{\pi}{6}$
- Identify all solutions in the given interval:
- $x = \frac{\pi}{6}$ (in the first quadrant)
- $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ (in the second quadrant)
For the equation $2 \sin 2x = 3 \cos x, 0° \leq x \leq 180°$:
- Rearrange: $2 \sin 2x - 3 \cos x = 0$
- Use the double angle formula: $2(2 \sin x \cos x) - 3 \cos x = 0$
- Factor out $\cos x$: $\cos x(4 \sin x - 3) = 0$
- Solve each factor:
$\cos x = 0 \to x = 90°$
$4\sin x - 3 = 0\to x = \arcsin\frac34 \to x\approx48.6°$
Therefore, the solutions are approximately 48.6° and 90°.
Don't forget to pay attention to the possible domain of $x$. For example, consider the following equation:
$\sin2x = \frac12, 0 \leq x \leq 2\pi$
If $0 \leq x \leq 2\pi$, that means $0 \leq 2x \leq 4\pi$! So the possible values of $2x$ are $\frac\pi6$, $\frac{5\pi}6$, $\frac{13\pi}6$, and $\frac{17\pi}6$, leading to:
$x = \frac{\pi}{12}$ or $x = \frac{5\pi}{12}$ or $x = \frac{13\pi}{12}$ or $x = \frac{17\pi}{12}$
If you see that $0 \leq x \leq \pi$ and assume that also applies to $2x$, you'll only get two of the solutions, and lose the final answer mark.
Equations with Multiple Angles
Equations involving multiple angles or combinations of trigonometric functions require careful manipulation before solving.
For $2 \tan(3(x-4)) = 1, -\pi \leq x \leq 3\pi$:
- Simplify: $\tan(3(x-4)) = \frac{1}{2}$
- Solve for the inner expression: $3(x-4) = \arctan(\frac{1}{2}) + n\pi$, where $n$ is an integer
- Solve for $x$: $x = \frac{\arctan(\frac{1}{2})}{3} + \frac{n\pi}{3} + 4$
- Find values of $n$ that give solutions in the interval $[-\pi, 3\pi]$
This yields multiple solutions within the given interval.
Quadratic Trigonometric Equations
Some trigonometric equations lead to quadratic equations in $\sin x$, $\cos x$, or $\tan x$. These are solved using standard quadratic equation techniques.
Quadratic in Sine or Cosine
For equations like $2 \sin^2 x + 5 \cos x + 1 = 0$ for $0 \leq x \leq 4\pi$:
- Substitute $u = \cos x$: $2(1-u^2) + 5u + 1 = 0$
- Simplify: $2u^2 - 5u - 3 = 0$
- Solve the quadratic in $u$
- Find $x = \arccos(u)$ for each solution of $u$
- Identify all solutions in the given interval
Students often forget to find all solutions in the given interval. Remember that $\arccos(u)$ gives only one solution, but there may be others in different quadrants.
Equations with Double Angles
Equations like $2 \sin x = \cos 2x, -\pi \leq x \leq \pi$ require the use of double angle formulas:
- Replace $\cos 2x$ with $1 - 2\sin^2 x$
- Rearrange to get a quadratic in $\sin x$: $4\sin^2 x + 2\sin x - 1 = 0$
- Solve this quadratic
- Find $x = \arcsin(u)$ for each solution
- Identify all solutions in $[-\pi, \pi]$
Graphical Method
Solving trigonometric equations graphically involves plotting the functions on both sides of the equation and finding their intersection points. This is usually done on paper 2, where you get a calculator to do this.
For example, to solve $0.866\sin(1.46x + 2.15) = 2.14\cos(0.559x - 1.23) - 0.291, 0 \leq x \leq 2\pi$, we can simply plot both sides:
In the graph above, the intersections of $y = 0.866\sin(1.46x + 2.15)$ and $y = 2.14\cos(0.559x - 1.23) - 0.291$ occur at $x \approx 0.157$ and $x \approx 4.00$. So, those are the solutions to our equation.
The graphical method is not viable if you don't have a calculator. So, use it on paper 2 if it's a one-mark "find solutions" question, or if you need to verify your analytical answers.
General Solutions
A general solution includes all possible solutions to a trigonometric equation, typically expressed using the variable $n$ to represent any integer.
For example, the general solution to $\sin x = \frac{1}{2}$ is:
$x = \frac{\pi}{6} + 2\pi n$ or $x = \pi - \frac{\pi}{6} + 2\pi n$, where $n$ is any integer.
