# How to Solve Trigonometric Equations - Part 1

By Amy Au - Head Maths Teacher and Co-founder at Pinnacle

Solving trigonometric equations is often one of the biggest struggles for students when progressing to senior maths. In this series of blogs, I hope to set out a basic framework for solving the most common types of trigonometric equations.

This is a multi-part series of blogs on how to solve trigonometric equations, covering content relevant to the NSW BOSTE Syllabus for Preliminary and HSC Mathematics and Extension 1 courses. Part 1 deals with the basics of solving a simple trigonometric equation, relevant to the Mathematics course.

Single Ratio Questions

This is the most basic type of equation with a single trigonometric ratio in the equation. All trigonometric equations, no matter how complicated, will need to be reduced down to a single ratio equation and solved having regard to the relevant quadrants and the reference angle.

Steps to solving a basic trigonometric equation:

1)      Make the trigonometric ratio the subject

2)      Identify the quadrants in which the solutions should be in by applying the ASTC rule

3)      Find the reference angle (this can be done with a calculator)

4)      Find all solutions in all relevant quadrants

Example 1 – Solve $$\sqrt 3 \tan x + 1 = 0$$  for $$0^\circ\le x \le 360^\circ$$

$$\begin{array}{l}\sqrt 3 \tan x + 1 = 0\\\tan x = - \frac{1}{{\sqrt 3 }}{\rm{ }}\\x = \left( {180^\circ- 30^\circ } \right),\left( {360^\circ- 30^\circ } \right)\\\therefore x = 150^\circ {\rm{, }}330^\circ \end{array}$$

Note: Since we are solving tan x to a negative ratio, the solutions are in the second and fourth quadrants where the tan ratio is negative. The reference angle is  $${\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = 30^\circ$$.

Example 2 – Solve $$4{\cos ^2}x - 1 = 0$$  for $$0^\circ\le x \le 360^\circ$$

$$\begin{array}{l}4{\cos ^2}x - 1 = 0\\{\cos ^2}x = \frac{1}{4}\\\cos x = \pm \frac{1}{2}\\x = 60^\circ ,\left( {180^\circ- 60^\circ } \right),\left( {180^\circ+ 60^\circ } \right),\left( {360^\circ- 60^\circ } \right)\\\therefore x = 60^\circ ,120^\circ ,240^\circ ,300^\circ \end{array}$$

Note: Since we are solving cos x to both positive and negative ratios, there is a solution in every quadrant. The reference angle is $${\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = 60^\circ$$ .

Single ratio question involving boundary angles

Boundary angles are referring to 0°, 90°, 180°, 270°, 360° … etc. A boundary angle solution occurs whenever sin x or cos x is solved to 0, 1 or -1, or when tan x is solved to 0. To find all solutions, one must observe the solutions from the graph rather than relying on quadrants and reference angles.

Example 3 – Solve $$\cos x = 0$$  for $$0^\circ\le x \le 360^\circ$$

As observed from the graph, $$\cos x$$  is 0 (i.e. at the x-axis) when $$x = 90^\circ$$ or $$270^\circ$$ .

Example 4 – Solve $${\mathop{\rm cosec}\nolimits} x = - 1$$ for $$0^\circ\le x \le 360^\circ$$

$$\begin{array}{l}{\mathop{\rm cosec}\nolimits} x = - 1\\\frac{1}{{\sin x}} = - 1\\\sin x = - 1\\\therefore x = 270^\circ \end{array}$$

Amy AuComment