Equations with multiple angles and solution intervals

The case of the angle multiplied

When the equation contains cos(2x) or sin(3x), we apply the same general formulas, but by substituting x with the full expression, then isolating x at the end.

Example: cos(2x) = cos(x)

We apply the formula directly: cos(2x) = cos(x) <=> 2x = x + 2kπ or 2x = -x + 2kπ.

First set: 2x – x = 2kπ, so x = 2kπ. Second set: 2x + x = 2kπ, so 3x = 2kπ, which gives x = 2kπ/3.

On [0, 2π[, the first family yields only x = 0. The second family yields, for k = 0, 1, 2: x = 0, 2π/3, 4π/3 (k = 3 would give 2π, which is outside the interval).

Final set (after removing duplicates): x = 0, 2π/3, 4π/3.

Method for counting solutions within an interval

1. Write the general solution with k an integer.
2. Substitute the values of k one by one (k = 0, 1, 2, … or k = -1, -2, …) as long as the result remains within the required interval.
3. Group all the values found into their respective families and remove duplicates.

Common pitfall number 1

Dividing both sides of an equation by cos(x) or sin(x): this can cause some solutions to disappear (those where cos(x) = 0, for example). It is always better to bring everything back to the form A = 0 and factorise.

Common pitfall number 2

Forgetting that an angle of the form 3x traverses the interval faster than x: if x traverses [0, 2π[, then 3x traverses [0, 6π[, so there are generally three times as many values of k to test.