Trigonometric Equations Revision Notes for HSC SSCE Mathematics Advanced

Trigonometric Equations

What is a trigonometric equation?

A trigonometric equation is an equation that includes trigonometric functions such as sine, cosine, or tangent. When you solve a trigonometric equation, you are finding the angles that make the equation true. These solutions are usually found within a specific restricted domain, such as $[0°, 360°]$ or $[0, 2\pi]$ .

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The restricted domain is typically chosen to capture one complete cycle of the trigonometric function. This ensures we find all unique solutions within a standard interval before considering periodicity for extended domains.

Solving basic trigonometric equations

Finding the principal angle

To solve equations like $\sin \theta = a$ , you use the inverse function $\sin^{-1}a$ to find the principal angle. The principal angle is the angle measured anticlockwise from the positive $x$ -axis, with a value in the interval $[0°, 360°]$ .

The general form is:

$\theta = \sin^{-1}a$

where:

$a$ is a constant where $-1 \leq a \leq 1$
$\theta$ is the angle that satisfies the equation

Using the CAST rule to find all solutions

The CAST rule (also called ASTC) helps you identify which trigonometric functions are positive in each quadrant of the unit circle. This is essential for finding all solutions to a trigonometric equation within the restricted domain.

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The CAST Rule:

Quadrant 1 (A): All trigonometric ratios are positive (cos $\theta$ , sin $\theta$ , tan $\theta$ all positive)
Quadrant 2 (S): Only sin $\theta$ is positive
Quadrant 3 (T): Only tan $\theta$ is positive
Quadrant 4 (C): Only cos $\theta$ is positive

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How to use the CAST rule:

When solving $\sin \theta = a$ :

If $a$ is positive, solutions are in Quadrants 1 and 2
If $a$ is negative, solutions are in Quadrants 3 and 4

Similar rules apply for $\cos \theta$ and $\tan \theta$ .

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Worked Example: Solving a basic equation

Problem: Solve $\cos \theta = 0.5$ for $\theta$ on $[0°, 360°]$ .

Solution:

Write the equation:

$\cos \theta = 0.5$

Take the inverse cosine of both sides:

$\theta = \cos^{-1}0.5$

Evaluate:

$\theta = 60°$

Since $0.5$ is positive, $\cos \theta$ is positive in Quadrants 1 and 4 (ASTC: A and C).

The reference angle is $60°$ (Quadrant 1).

In Quadrant 4, the angle is:

$360° - 60° = 300°$

Therefore, the solutions are $\theta = 60°, 300°$ .

Extended domain equations

Understanding periodicity

Trigonometric functions are periodic, meaning they repeat their values at regular intervals:

$\sin \theta$ and $\cos \theta$ repeat every $360°$ (or $2\pi$ radians)
$\tan \theta$ repeats every $180°$ (or $\pi$ radians)

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The periodic nature of trigonometric functions means that if $\theta$ is a solution, then $\theta + 360°$ (or $\theta + 2\pi$ ) is also a solution for sine and cosine. For tangent, $\theta + 180°$ (or $\theta + \pi$ ) is also a solution.

Solving equations in extended domains

To solve equations in domains beyond $[0°, 360°]$ , such as $[0°, 720°]$ or $[-180°, 180°]$ :

First find all solutions in $[0°, 360°]$
Then add or subtract multiples of $360°$ (or $2\pi$ ) to cover the extended domain

This process uses the periodic nature of trigonometric functions to identify all possible angles that satisfy the equation.

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Worked Example: Extended domain

Problem: Solve $\tan \theta = 1$ for $\theta$ on $[0, 4\pi]$ .

Solution:

Write the equation:

$\tan \theta = 1$

Take the inverse tangent of both sides:

$\theta = \tan^{-1}1$

Evaluate:

$\theta = \frac{\pi}{4}$

Since $1$ is positive, $\tan \theta$ is positive in Quadrants 1 and 3 (ASTC: A and T).

In $[0, 2\pi]$ , solutions are $\frac{\pi}{4}$ (Quadrant 1) and $\frac{\pi}{4} + \pi = \frac{5\pi}{4}$ (Quadrant 3).

Now add $2\pi$ to each solution to cover $[0, 4\pi]$ :

$\theta = \frac{\pi}{4} + 2\pi$

$= \frac{9\pi}{4}$

$\theta = \frac{5\pi}{4} + 2\pi$

$= \frac{13\pi}{4}$

The solutions are $\theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}$ .

Quadratic trigonometric equations

What are quadratic trigonometric equations?

Some trigonometric equations can be rearranged into quadratic form in terms of a trigonometric function. For example, an equation like $2\sin^2\theta - \sin\theta - 1 = 0$ is quadratic in $\sin\theta$ .

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Method for solving quadratic trigonometric equations:

Substitute a variable (e.g., $u = \sin\theta$ ) to convert to a standard quadratic equation
Solve the quadratic using factorisation or the quadratic formula
Find angles for each value of the variable using inverse functions and the ASTC mnemonic
Ensure all angles are within the restricted domain

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Worked Example: Quadratic equation

Problem: Solve $2\sin^2\theta - \sin\theta - 1 = 0$ for $\theta$ on $[0°, 360°]$ .

Solution:

Write the equation:

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

Substitute $u = \sin\theta$ :

$2u^2 - u - 1 = 0$

Using the quadratic formula to solve for $u$ :

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$u = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 2 \times (-1)}}{2 \times 2}$

$= \frac{1 \pm \sqrt{1 + 8}}{4}$

$= \frac{1 \pm \sqrt{9}}{4}$

$= \frac{1 \pm 3}{4}$

$= 1, -\frac{1}{2}$

Since $u = \sin\theta$ , solve for $\theta$ where $-1 \leq u \leq 1$ .

For $\sin\theta = 1$ :

$\sin\theta = 1$

$\theta = \sin^{-1}1$

$= 90°$

For $\sin\theta = -\frac{1}{2}$ :

$\sin\theta = -\frac{1}{2}$

$\theta = \sin^{-1}\left(-\frac{1}{2}\right)$

$= 210°, 330°$

Since $-\frac{1}{2}$ is negative, we take the angles in Quadrants 3 and 4 (where sine is negative).

The solutions are $\theta = 90°, 210°, 330°$ .

Checking your solutions

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Always verify your answers by substituting each value of $\theta$ back into the original equation. This helps catch any errors and confirms that all solutions satisfy the equation.

For example, checking $\theta = 90°$ in $2\sin^2\theta - \sin\theta - 1$ :

$\text{LHS} = 2\sin^2 90° - \sin 90° - 1$

$= 2 \times 1^2 - 1 - 1$

$= 2 - 1 - 1$

$= 0$

$= \text{RHS}$

The solution is correct.

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Key Points to Remember:

Inverse functions help you find the principal angle, which is your starting point for finding all solutions
The CAST rule tells you which trigonometric functions are positive in each quadrant, helping you identify where all solutions lie
Periodicity allows you to extend solutions beyond $[0°, 360°]$ by adding or subtracting multiples of the period ( $360°$ or $2\pi$ for sine and cosine, $180°$ or $\pi$ for tangent)
Quadratic trigonometric equations require substitution to convert them into standard quadratic form before solving
Always verify your solutions by substituting them back into the original equation

Trigonometric Equations (HSC SSCE Mathematics Advanced): Revision Notes