If you get the first answer right and still lose marks, the problem is usually counting the interval, not solving the equation. A calculator gives one angle. The interval in the question may hold two, four or more, and each one needs to be found.
This page shows a method that keeps every valid solution and explains the two habits that lose them. It supports trigonometric equations and graphs.
Why are solutions lost?
There are three common causes:
- Stopping at the principal value. The calculator’s answer is one of the solutions.
- Forgetting to change the interval when the angle is 2x, 3x or x + 30°.
- Dividing by a trig term that can be zero, which removes solutions.
A method that keeps every solution
- Write the interval for the whole angle. If the question says 0° ≤ x ≤ 360° and the equation has 2x, then the interval for 2x is 0° ≤ 2x ≤ 720°.
- Find the principal value with the inverse function.
- Find the related angles using a sketch of the graph or the quadrant diagram.
- Add full turns while the angle stays inside the interval.
- Convert back: divide by 2, or subtract 30°, to get x.
- Check by substituting each answer into the original equation.
Worked example
Solve sin 2x = 0.5 for 0° ≤ x ≤ 360°.
Step 1, interval for 2x. Multiply the interval by 2: 0° ≤ 2x ≤ 720°.
Step 2, principal value. sin⁻¹(0.5) = 30°.
Step 3, related angle. Sine is also positive in the second quadrant: 180° − 30° = 150°.
Step 4, add full turns. 30° + 360° = 390°, and 150° + 360° = 510°. Both are below 720°. The next turn would give 750° and 870°, which are outside. So 2x = 30°, 150°, 390°, 510°.
Step 5, halve. x = 15°, 75°, 195°, 255°.
Step 6, check. sin 30° = 0.5. sin 150° = 0.5. sin 390° = sin 30° = 0.5. sin 510° = sin 150° = 0.5. All four agree, so x = 15°, 75°, 195°, 255°.
A student who stops at step 2 writes x = 15°, and one who skips step 1 finds only 15° and 75°. Both lose marks on the same equation.
The divide-by-sin trap
A second way to lose answers: solve 2 sin²x = sin x for 0° ≤ x ≤ 360°.
Mistaken working: divide by sin x, giving 2 sin x = 1, so sin x = 1/2 and x = 30°, 150°.
This loses x = 0°, 180° and 360°, because sin x = 0 there.
The correction is to move everything to one side and factorise: 2 sin²x − sin x = 0, so sin x (2 sin x − 1) = 0. Then either sin x = 0, giving x = 0°, 180°, 360°, or sin x = 1/2, giving x = 30°, 150°. There are five solutions in total: 0°, 30°, 150°, 180°, 360°.
Check yourself
1. Solve cos 3x = 0.5 for 0° ≤ x ≤ 180°.
Show answer
The interval for 3x is 0° ≤ 3x ≤ 540°. cos⁻¹(0.5) = 60°. The other angle in a turn is 360° − 60° = 300°. Adding 360° to 60° gives 420°, which is allowed. Adding 360° to 300° gives 660°, which is too big.
So 3x = 60°, 300°, 420°, and x = 20°, 100°, 140°. Check: cos 60° = cos 300° = cos 420° = 0.5.
2. Solve sin(x + 30°) = −0.5 for 0° ≤ x ≤ 360°.
Show answer
The interval for x + 30° is 30° ≤ x + 30° ≤ 390°. Sine is −0.5 at 210° and 330° within one turn. Both are in the interval, and adding 360° goes beyond 390°.
So x + 30° = 210° or 330°, and x = 180° or 300°. Check: sin 210° = −0.5 and sin 330° = −0.5.
3. Solve tan x = 1 for 0° ≤ x ≤ 360°. Explain in a sentence why there are two answers.
Show answer
tan⁻¹(1) = 45°, and tangent repeats every 180°, so the next solution is 225°. The answers are x = 45° and x = 225°. The second lies in the third quadrant, where tangent is positive again.
What can you do next?
Use the triangle and bearings reasoning board to build the habit of stating the interval and angle units first. The non-calculator working trainer supports exact values such as sin 30° = 1/2.
Then test yourself with the original practice hub. If you keep counting one answer short even when you know the method, online one-to-one Additional Mathematics tuition gives you an assigned teacher. They can watch the exact step where the missing solutions slip away.