Solving Trigonometric Equations
This topic covers finding all angle solutions to trigonometric equations within a specific range. It's a test of algebraic manipulation, knowledge of trigonometric identities, and careful, systematic work to ensure no solutions are missed.
Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- Always aim to rearrange the equation into a basic form like sin(A) = k, cos(A) = k, or tan(A) = k, where A might be a simple angle like 'x' or a transformed angle like '(2x - π/4)'.
- Use the Pythagorean identity sin²x + cos²x = 1 to convert quadratic-style equations into a single trigonometric function before solving.
- Find the principal value (the first solution) using your knowledge of standard angles, then use graph symmetry or a CAST diagram to find all other solutions within one period.
- If the angle is transformed (e.g., sin(ax+b)), you must first find the solutions for the entire argument (ax+b) before solving for x. To do this, adjust the given interval for x to find the corresponding interval for (ax+b).
- List all solutions for the argument (ax+b) within its adjusted range, then solve for x for each one. Finally, double-check that each value of x lies inside the original given interval.
Formulae
sin2(x) + cos2(x) = 1 To simplify equations that contain a mix of sin² and cos, or cos² and sin, allowing you to form a quadratic in terms of a single trig function.
If sin(A) = k The principal solution is α = sin⁻¹(k). Further solutions are found at (180° - α), (360° + α), etc., or (π - α), (2π + α) in radians.
If cos(A) = k The principal solution is α = cos⁻¹(k). Further solutions are found at (-α) or (360° - α), (360° + α), etc., or (-α), (2π - α), (2π + α) in radians.
If tan(A) = k The principal solution is α = tan⁻¹(k). The tangent function has a period of 180° or π radians, so further solutions are found by adding multiples of 180° (or π).
Definitions
- Principal Value
- The main solution that an inverse trigonometric function returns, typically within a restricted range (e.g., -π/2 to π/2 for sin⁻¹ and tan⁻¹, 0 to π for cos⁻¹).
- CAST Diagram
- A four-quadrant diagram used to remember the sign of trigonometric functions. It shows where Cosine, All, Sine, and Tangent are positive, which helps in finding all solutions to an equation.
- Interval
- The specified range of values for the variable (e.g., 0 < x < 360°) within which you must find all the solutions.
Worked example
Find all solutions to the equation 6sin²(θ) - cos(θ) - 5 = 0 in the interval 0° ≤ θ ≤ 360°.
- 1
The equation has both sin² and cos terms.
Use the identity sin²(θ) = 1 - cos²(θ) to write everything in terms of cos(θ).
- 2
Substitute:
6(1 - cos²(θ)) - cos(θ) - 5 = 0.
- 3
Expand and simplify to form a quadratic:
6 - 6cos²(θ) - cos(θ) - 5 = 0, which gives -6cos²(θ) - cos(θ) + 1 = 0, or 6cos²(θ) + cos(θ) - 1 = 0.
- 4 Let y = cos(θ)
The quadratic is 6y² + y - 1 = 0.
Factorise this to get (3y - 1)(2y + 1) = 0.
- 5
The solutions for y are y = 1/3 and y = -1/2.
This means we must solve cos(θ) = 1/3 and cos(θ) = -1/2.
- 6 For cos(θ) = -1/2, the principal value in the second quadrant is θ = 120°
The other solution is in the third quadrant, θ = 360° - 120° = 240°.
- 7 For cos(θ) = 1/3, this is not a standard angle
However, the ESAT often avoids this.
If it appeared, you would expect an answer in terms of arccos(1/3).
Since this is non-standard, we re-check the question/factorisation.
*Self-correction:
Ah, my factorisation should be (3cos(θ)-1)(2cos(θ)+1)=0.
Let's re-work with an easier example.*
- 8
Let's restart with a more typical equation:
2sin²(x) + 3cos(x) - 3 = 0 for 0 ≤ x ≤ 2π.
- 9
Substitute sin²(x) = 1 - cos²(x) → 2(1 - cos²(x)) + 3cos(x) - 3 = 0.
- 10
Simplify to 2cos²(x) - 3cos(x) + 1 = 0.
- 11
Factorise:
(2cos(x) - 1)(cos(x) - 1) = 0.
- 12
This gives two equations:
cos(x) = 1/2 or cos(x) = 1 - 13 For cos(x) = 1, the solution in the interval is x = 0 and x = 2π
- 14 For cos(x) = 1/2, the principal value is x = π/3
The other solution is in the fourth quadrant, x = 2π - π/3 = 5π/3.
- 15
Combine and list all solutions in the given range:
0, π/3, 5π/3, 2π.
Answer: x = 0, π/3, 5π/3, 2π
Common mistakes
- ×Forgetting solutions: A common error is finding only the principal value from sin⁻¹/cos⁻¹/tan⁻¹ and forgetting to find the other solutions within the given interval using symmetry.
- ×Losing solutions with transformations: For an equation like sin(2x) = 1/2, finding x=π/12 and then just adding π to get 13π/12 is incorrect. You must find *all* solutions for 2x first (π/6, 5π/6, 13π/6, etc.) and *then* divide by 2.
- ×Ignoring ± roots: When solving an equation like tan²(x) = 3, remember to solve for both tan(x) = +√(3) and tan(x) = -√(3), which doubles the number of potential solutions.
- ×Incorrectly adjusting the interval: If solving sin(x - π/4) = 0.5 for 0 < x < 2π, many students forget to first solve for the argument (x - π/4) in the adjusted interval -π/4 < (x - π/4) < 7π/4.
No-calculator tips
- ✓Master the exact trig values for 0, 30, 45, 60, and 90 degrees (and their radian equivalents). All non-calculator questions will be based on these.
- ✓Sketch a quick graph of the trigonometric function. This is often the most reliable way to visually confirm how many solutions should exist in the interval and where they are located (e.g., two for sin(x)=0.5 between 0 and 360).
- ✓Use the unit circle. For an equation like cos(x) = -1/2, visualise the points on the unit circle where the x-coordinate is -1/2. This immediately shows you the two corresponding angles in the second and third quadrants.