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CIE A-Level Maths Study Notes

1.5.5 Solving Trigonometric Equations

In this section, we delve into the intricate world of solving trigonometric equations, a pivotal component of A-Level Pure Mathematics. Mastery of these techniques is essential for excelling in various mathematical challenges encountered in the curriculum.

Introduction

Solving trigonometric equations effectively hinges on understanding and applying various mathematical strategies. Here's a streamlined approach:

1. Isolate the Trigonometric Function: Move the trigonometric term (e.g., sinsin, coscos, tantan) to one side to simplify the equation.

2. Use Trigonometric Identities: Apply identities like the Pythagorean identity and angle sum/difference identities to simplify the equation into a more solvable form.

3. Acknowledge Periodicity: Understand that trigonometric functions repeat values at regular intervals, which is key to identifying all solution possibilities.

4. Apply Inverse Functions: Utilize inverse trigonometric functions to find the principal solution, serving as a starting point for all possible solutions.

5. Identify All Solutions: Considering the trigonometric function's periodicity, find all solutions within a given interval, ensuring a comprehensive solution set.

Examples

Example 1: 3sinx+1=03\sin x + 1 = 0

Solution:

1. Isolate sinx\sin x:

3sinx=13\sin x = -1

sinx=13\sin x = -\frac{1}{3}

2. Principal Value:

x=arcsin(13)x = \arcsin\left(-\frac{1}{3}\right)

The principal value is approximately 0.34-0.34 radians.

3. General Solutions:

Given the periodic nature of the sine function:

  • First form: x=arcsin(13)+2nπx = \arcsin\left(-\frac{1}{3}\right) + 2n\pi
  • Second form: x=πarcsin(13)+2nπx = \pi - \arcsin\left(-\frac{1}{3}\right) + 2n\pi

Where nn is any integer, accounting for the sine function's periodicity of 2π2\pi.

Example 2: 3sinx5cosx=13\sin x - 5\cos x = 1

Solution:

1. Trigonometric Identities Substitution:

sinx=2tan(x2)1+tan2(x2)\sin x = \frac{2\tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)},

cosx=1tan2(x2)1+tan2(x2)\cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}

2. Substitute and Simplify:

Substitute these into 3sinx5cosx=13\sin x - 5\cos x = 1, resulting in a simplified equation in terms of tan(x2)\tan\left(\frac{x}{2}\right).

3. Solve for tan(x2)\tan\left(\frac{x}{2}\right):

Solutions for tan(x2)\tan\left(\frac{x}{2}\right) are 34+334-\frac{3}{4} + \frac{\sqrt{33}}{4} and 34334-\frac{3}{4} - \frac{\sqrt{33}}{4}.

4. Find xx Using Inverse Tangent:

Calculate xx values: x=2arctan(34±334)x = 2\arctan\left(-\frac{3}{4} \pm \frac{\sqrt{33}}{4}\right)

5. Periodicity and General Solutions:

Considering the periodicity of tan(x2)\tan\left(\frac{x}{2}\right), which leads to a periodicity of 2π2\pi for xx, the general solutions for xx include adding multiples of 2π2\pi to each solution.

General Solutions:

  • x=2arctan(34334)+2nπx = -2\arctan\left(\frac{3}{4} - \frac{\sqrt{33}}{4}\right) + 2n\pi and
  • x=2arctan(34+334)+2nπx = -2\arctan\left(\frac{3}{4} + \frac{\sqrt{33}}{4}\right) + 2n\pi, for nn being any integer.
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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