# Solution of simple trigonometric equations

> Edexcel International GCSE Further Pure Mathematics · 4PM1
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s10-solution-of-simple-trigonometric-equations/

This guide teaches you to solve linear and quadratic trigonometric equations for Edexcel IGCSE Further Pure Maths (4PM1). You will learn to handle shifted/multiple angles, adjust intervals, and find all valid solutions in the given range.

**Prerequisites:** [Trigonometric identities (Pythagorean, double angle)](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s10-trigonometric-identities/); [Exact trigonometric values for 30°, 45°, 60°](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s10-exact-trig-values/)

## Learning objectives

- Solve linear trigonometric equations with multiple/shifted angles over a given interval
- Solve quadratic trigonometric equations using the Pythagorean identity
- Adjust intervals for transformed angles and unwind to find all valid x solutions

## Linear trigonometric equations with single angles

Linear trigonometric equations are of the form $\sin x = k$, $\cos x = k$ or $\tan x = k$, where $k$ is a constant. Always start by noting the unit of the given interval (degrees or radians) and use it consistently for all working.

**Principal solution** — The smallest positive solution to a trigonometric equation, found using your calculator or exact value knowledge. Use quadrant rules to find all other solutions in the interval.

**Worked example:** Solve $\tan x = 1$ for $90^\circ < x < 270^\circ$.

1. Find the principal solution using exact value knowledge: $\tan^{-1}(1) = 45^\circ$
2. Tangent repeats every 180°, so add 180° to the principal solution to find values in the given range:
3. $$x = 45^\circ + 180^\circ = 225^\circ$$
4. Verify 225° is between 90° and 270°, so it is the only valid solution.

> **Exam tip:** Always test each solution you find by plugging it back into the original equation to confirm it works, and that it lies inside the given interval.

*Calculator:* allowed

## Working with transformed (shifted/multiple) angles

When the argument of the trigonometric function is a linear transformation of x (e.g. $3x + 30^\circ$ or $x - \pi/2$), first adjust the given interval to match the transformed variable, find all solutions for the transformed variable, then rearrange to solve for x.

**Worked example:** Solve $\cos(3x + 30^\circ) = \frac{1}{2}$ for $-90^\circ < x < 90^\circ$.

1. Adjust interval for $\theta = 3x + 30^\circ$: multiply x bounds by 3, add 30°: $3(-90) +30 = -240^\circ$, $3(90) +30 = 300^\circ$. Solve $\cos \theta = 1/2$ for $-240^\circ < \theta < 300^\circ$.
2. Principal solution: $\cos^{-1}(1/2) = 60^\circ$. Cosine is positive in first and fourth quadrants, so valid $\theta$ values are $-60^\circ$ and $60^\circ$.
3. Unwind to x: $x = \frac{\theta - 30^\circ}{3}$
4. $$\text{For } \theta = -60^\circ: x = \frac{-90}{3} = -30^\circ$$
5. $$\text{For } \theta = 60^\circ: x = \frac{30}{3} = 10^\circ$$
6. Both solutions are within $-90^\circ < x < 90^\circ$, so final solutions are $x=-30^\circ, 10^\circ$.

> **Exam tip:** Always widen the interval for the transformed angle first before solving, otherwise you will miss solutions that fall outside the original x interval before unwinding.

*Calculator:* allowed

## Solving quadratic trigonometric equations

Quadratic trigonometric equations can be rearranged to the form $aR^2 + bR + c = 0$ where R is a single trigonometric ratio (sin, cos, tan). Use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to eliminate one ratio if the equation has both sin and cos terms.

**Pythagorean trigonometric identity** — $\sin^2 x + \cos^2 x = 1$ for all values of x, used to convert between squared sine and cosine terms to reduce an equation to a single trigonometric ratio.

**Worked example:** Solve $6\cos^2 x^\circ + \sin x^\circ - 5 = 0$ for $0 ≤ x < 360$.

1. Replace $\cos^2 x$ with $1 - \sin^2 x$ using the Pythagorean identity:
2. $$6(1 - \sin^2 x) + \sin x - 5 = 0$$
3. $$6 - 6\sin^2 x + \sin x - 5 = 0 \implies 6\sin^2 x - \sin x - 1 = 0$$
4. Factorise the quadratic in $\sin x$:
5. $$(3\sin x + 1)(2\sin x - 1) = 0$$
6. Solve each factor: $\sin x = -\frac{1}{3}$ or $\sin x = \frac{1}{2}$
7. Find all solutions in range: For $\sin x = 1/2$: $30^\circ, 150^\circ$. For $\sin x = -1/3$: $199.5^\circ, 340.5^\circ$ (rounded to 1 decimal place).

> **Exam tip:** Discard any quadratic solutions where $|\sin x|>1$ or $|\cos x|>1$, as these are impossible and have no real solutions.

*Calculator:* allowed

## Solving equations in radians

Many questions use radians instead of degrees, especially for intervals involving $\pi$. Ensure your calculator is set to radians mode for these questions, and adjust intervals using $2\pi$ for full rotations instead of 360°.

**Worked example:** Solve $\sin(x - \pi/2) = 3/4$ for $0 < x < 2\pi$.

1. Adjust interval for $\theta = x - \pi/2$: subtract $\pi/2$ from bounds: $-\pi/2 < \theta < 3\pi/2$.
2. Principal solution: $\sin^{-1}(3/4) ≈ 0.848$ radians. Sine is positive in first and second quadrants, so second solution is $\pi - 0.848 ≈ 2.294$ radians.
3. Unwind to x: $x = \theta + \pi/2$
4. $$\text{First solution: } 0.848 + 1.571 ≈ 2.42 \text{ radians}$$
5. $$\text{Second solution: } 2.294 + 1.571 ≈ 3.87 \text{ radians}$$
6. Both values are within $0 < x < 2\pi$, so they are valid.

> **Exam tip:** If the question asks for exact solutions, use exact radian values for $30^\circ=\pi/6$, $45^\circ=\pi/4$, $60^\circ=\pi/3$ etc. instead of decimal approximations.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Mixing degrees and radians in the same working
  - Why it fails: Causes incorrect solution values as calculator modes will give wrong outputs
  - Correct: Always use the unit stated in the question interval, set calculator to match, and keep all working in that unit
- **Wrong:** Forgetting to adjust the interval for transformed angles
  - Why it fails: You will miss valid solutions that fall outside the original x interval when solving for the transformed variable
  - Correct: First widen/narrow the interval to match the transformed angle before finding solutions
- **Wrong:** Including general solutions with $+360n^\circ$ or $+2\pi n$
  - Why it fails: The exam only requires solutions in the given interval, extra general solutions will lose marks
  - Correct: Only list solutions that lie strictly inside the given interval, discard any outside
- **Wrong:** Not discarding quadratic trigonometric solutions where $|\sin x|>1$ or $|\cos x|>1$
  - Why it fails: Sine and cosine only take values between -1 and 1, so these solutions are impossible
  - Correct: After solving the quadratic, check each root is within the valid range for the trigonometric ratio before finding x values
- **Wrong:** Rounding solutions too early in working
  - Why it fails: Rounding intermediate steps leads to inaccurate final answers
  - Correct: Keep 3-4 extra decimal places in intermediate steps, only round final answers to the required number of significant figures or decimal places

## Cheatsheet

| Equation type | Steps to solve | Key check |
| --- | --- | --- |
| Linear (single angle) | 1. Find principal solution 2. Find all solutions in interval 3. Verify each solution | Confirm all solutions are within given interval |
| Linear (transformed angle) | 1. Adjust interval for transformed variable 2. Solve for transformed variable 3. Unwind to x 4. Verify | Widen interval before solving, don't miss solutions |
| Quadratic | 1. Use Pythagorean identity to get single ratio 2. Factorise/solve quadratic 3. Discard invalid roots 4. Find all x solutions in interval | Check $\|\sin x\| ≤1$ and $\|\cos x\| ≤1$ for all roots |

## What's next

Now that you can solve simple trigonometric equations, you are ready to apply these skills to more advanced trigonometry problems in your Edexcel IGCSE Further Pure Maths exam. Next, you should practice solving mixed trigonometric equation questions from past papers, and learn how to use double-angle identities to solve more complex equations. You will also use these solution skills when working with trigonometric graphs and calculus problems later in the course.

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