1The idea in one paragraph

An equation like sin x = ½ asks where the sine wave is at height ½. Because the wave repeats, the answer is not one angle but a list of them, one or two in every turn, and the question fixes which ones you want by giving an interval. So solving has three steps: find one angle (the principal value, from exact values or from sin⁻¹), use the symmetry of the unit circle to find the second angle in the same turn, then add or subtract whole periods to collect every solution inside the interval. Harder equations are turned into this simple form first: an identity removes a double angle or a squared ratio, the result is a quadratic in one ratio, and each root of that quadratic is a simple equation again. On Paper 2 you can skip the algebra and read the solutions off a graph.

3The basic method

The principal value is the one angle the inverse function gives: sin⁻¹ and tan⁻¹ return an angle between −π/2 and π/2, cos⁻¹ an angle between 0 and π. The second angle in the same turn comes from the symmetries of 3.5, which Figure 2 shows on the circle.

sin x = k: x and π − x. cos x = k: x and −x (or 2π − x). tan x = k: x and π + x. Then add or subtract 2π (π for tan) until you leave the interval.

Worked example 1 (Paper 1). Solve 2 cos x + √3 = 0 for 0 ≤ x ≤ 2π.

Here the related-angle method from 3.5 is quicker than a principal value, and on Paper 1 it is the method to use: find the acute angle for the size, then place it in each quadrant where the ratio has the right sign.

Worked example 2 (Paper 2, degrees). Solve tan x = −2.5 for 0° ≤ x ≤ 360°.

Worked example 3 (Paper 2, a negative interval). Solve sin x = −0.4 for −π ≤ x ≤ π.

A quick sketch of the wave on the interval tells you how many solutions to expect before you start. Here sin x is negative only for −π < x < 0 in the interval, and dips to −1 once, so there are exactly two.

4When the angle is transformed

In sin 2x = k or cos(3(x − 4)) = k the unknown inside the function is not x. The clean method is a substitution that you write down.

1. Let u be the whole inside of the function: u = 2x.
2. Transform the interval: if 0 ≤ x ≤ 2π then 0 ≤ u ≤ 4π.
3. Solve for u in the new interval, collecting every solution.
4. Convert each back to x.

Skipping step 2 is the commonest way to lose solutions: sin 2x completes two waves in 0 ≤ x ≤ 2π, so there are twice as many answers as for sin x.

Worked example 4 (Paper 1). Solve 2 sin 2x = √3 for 0 ≤ x ≤ 2π.

Figure 3 confirms four solutions.

Worked example 5 (a translated angle). Solve sin(2(x + π/6)) = −½ for 0 ≤ x ≤ π.

Check the first: 2(5π/12 + π/6) = 2 × 7π/12 = 7π/6, and sin(7π/6) = −½ ✓.

5Solving graphically on the GDC

On Paper 2 any equation can be solved by graphing each side and finding the intersections. It is the guide's "graphically", and it is often the fastest route. Three habits make it reliable.

• Check the mode. Radians unless the question uses degrees.
• Set the window to the interval. Solutions outside it are not wanted, and solutions just inside it are easy to miss with a default window.
• Count before you read. Look at the whole interval and count the crossings; then find each one.

Worked example 6. Solve 2 tan(2(x − 1)) = 3 for 0 ≤ x ≤ 4.

Graph y = 2 tan(2(x − 1)) and y = 3 with 0 ≤ x ≤ 4, as in Figure 4. The tangent has period π/2 ≈ 1.57, so the interval holds between two and three branches. The line y = 3 crosses two of them: x = 1.49 and x = 3.06. The near-vertical lines the calculator may draw at the asymptotes are not part of the graph and are not solutions.

The same equation by algebra: tan(2(x − 1)) = 1.5, so 2(x − 1) = 0.983 or 0.983 + π = 4.12, giving x = 1.49 and 3.06. Both methods agree; on Paper 2 either earns full marks if the working or a sketch is shown.

6Equations that need an identity first

When an equation contains two different ratios, or a ratio and a double angle, it cannot be solved as it stands. The plan, summarised in Figure 5, is always to rewrite it in one ratio of one angle. That usually produces a quadratic.

Worked example 7 (Pythagorean identity to a quadratic). Solve 2 cos²x + 3 sin x − 3 = 0 for 0 ≤ x ≤ 2π.

If the factorisation is not obvious, write s = sin x and solve 2s² − 3s + 1 = 0 by the quadratic formula. The letter makes the structure visible.

Worked example 8 (double angle to a quadratic, with a rejection). Solve cos 2x + 5 cos x + 3 = 0 for 0 ≤ x ≤ 2π.

The rejection needs its reason written down: "no solution, as −1 ≤ cos x ≤ 1". An examiner gives an R1 or A1 for it.

Worked example 9 (double angle to factorising; the dividing trap). Solve sin 2x = sin x for 0 ≤ x ≤ 2π.

Dividing both sides by sin x in the second line would leave cos x = ½ and only two answers. Figure 6 shows the other three are real solutions: the curves meet five times. Never divide by an expression that could be zero; factorise instead.

Worked example 10 (when dividing is safe). Solve sin x = √3 cos x for 0 ≤ x ≤ 2π.

If cos x = 0 then sin x = ±1, and ±1 = √3 × 0 is false, so no solution has cos x = 0. Dividing by cos x is therefore safe, and gives tan x = √3, so x = π/3 or π/3 + π = 4π/3. The one-line check that cos x = 0 is not a solution is what justifies the division; write it.

A quadratic in tan. tan²x = 3 for 0 ≤ x ≤ π gives tan x = √3 or tan x = −√3, so x = π/3 or x = 2π/3. Keep both square roots: the negative one is as real as the positive.