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CIE A-Level Maths Study Notes

2.3.3 Solving Trigonometric Equations

Trigonometric equations are vital for understanding numerous mathematical and scientific concepts. This guide is tailored for students, providing comprehensive strategies and solutions for solving trigonometric equations.

Introduction to Trigonometric Equations

These equations involve trigonometric functions (sine, cosine, tangent, and their reciprocals) set equal to a value. The goal is to find all angles (usually represented as tt or θ\theta) that satisfy the equation.

Strategies for Solving Trigonometric Equations

1. Identifying the Function: Ascertain which of the six primary trigonometric functions are involved.

2. Isolating the Trigonometric Function: Employ algebraic manipulation to isolate the function on one side of the equation.

3. Using Inverse Trigonometric Functions: Apply inverse functions to solve for the angle.

4. Considering All Possible Angles: Be aware of the periodic nature of these functions.

5. Utilising Trigonometric Identities: Use identities to simplify and solve equations.

Methods to Isolate and Solve for Unknown Angles

Algebraic Manipulation: Techniques include expanding, factoring, or simplifying expressions.

Graphical Interpretation: Understanding the periodicity and symmetry of trigonometric functions through graphs.

Substitution: Use identities like sin2(t)+cos2(t)=1\sin^2(t) + \cos^2(t) = 1 for simplification.

Example Problems

Example 1:

Solve tan(t)+cot(t)=4\tan(t) + \cot(t) = 4:

Solution:

1. Rewrite cot(t)\cot(t) as 1tan(t)\frac{1}{\tan(t)}: tan(t)+1tan(t)=4\tan(t) + \frac{1}{\tan(t)} = 4.

2. Clear the fraction by multiplying by:

3. Form a quadratic: tan2(t)4tan(t)+1=0\tan^2(t) - 4\tan(t) + 1 = 0.

4. Use quadratic formula: solutions are t=π12,5π12t = \frac{\pi}{12}, \frac{5\pi}{12}.

5. Account for periodicity: general solutions are t=π12+πkt = \frac{\pi}{12} + \pi k and t=5π12+πkt = \frac{5\pi}{12} + \pi k, where kk is an integer.

Example 2:

Solve 2sec(t)5tan(t)=22\sec(t) - 5\tan(t) = 2:

Solution:

1. Use sec(t)=1cos(t)\sec(t) = \frac{1}{\cos(t)}: 2cos(t)5tan(t)=2\frac{2}{\cos(t)} - 5\tan(t) = 2.

2. Multiply bycos(t) \cos(t): 25sin(t)=2cos(t)2 - 5\sin(t) = 2\cos(t).

3. Rearrange: 2cos(t)+5sin(t)=22\cos(t) + 5\sin(t) = 2.

4. Square and simplify using sin2(t)+cos2(t)=1\sin^2(t) + \cos^2(t) = 1.

5. Identify that the equation holds for all tt, with specific solutions at t=2πkt = 2\pi k and t=πarctan(2021)+2πkt = \pi - \arctan\left(\frac{20}{21}\right) + 2\pi k, where kk is an integer.

Example 3:

Solve 3cos2(t)+sin(t)=13\cos^2(t) + \sin(t) = 1:

Solution:

1. Use identity cos2(t)=1sin2(t)):3(1sin2(t))+sin(t)=1\cos^2(t) = 1 - \sin^2(t) ): 3(1 - \sin^2(t)) + \sin(t) = 1.

2. Form quadratic in sin(t)):3sin2(t)sin(t)2=0\sin(t) ): 3\sin^2(t) - \sin(t) - 2 = 0.

3. Solve for sin(t)\sin(t) using quadratic formula.

4. Find tt using sin1\sin^{-1}.

5. Account for periodicity: general solutions are t=π2+2πkt = \frac{\pi}{2} + 2\pi k, t=arctan(25)+2πkt = -\arctan\left(\frac{2}{\sqrt{5}}\right) + 2\pi k, and t=π+arctan(25)+2πkt = -\pi + \arctan\left(\frac{2}{\sqrt{5}}\right) + 2\pi k, where kk is an 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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