To solve a trig equation on an interval, find one basic angle from the calculator, use the sign to pick the quadrants where the answer lives, and write down every angle inside the interval. This appears whenever a question says “solve for 0° ≤ x ≤ 360°”.
The skill is the starting point of trigonometric equations and graphs. It depends on the exact values and the graph shapes you have met before, and it feeds every later lesson in this module.
How do you build every solution from one angle?
Start by isolating the trig function, so the equation reads sin x = k, cos x = k or tan x = k. Then follow these steps.
- Find the basic angle. Ignore the sign of k and work out the inverse function of its size. This gives an acute angle, call it α.
- Decide the quadrants. Use the sign of k. Sine is positive in quadrants 1 and 2, cosine in 1 and 4, and tangent in 1 and 3.
- Build the angles. Quadrant 1 gives α, quadrant 2 gives 180° − α, quadrant 3 gives 180° + α and quadrant 4 gives 360° − α.
- Keep only those inside the interval. Add or subtract 360° if needed, and stop when you leave the stated range.
For sin and cos the two answers per cycle are symmetrical about a line, and for tan they are exactly 180° apart.
Worked example
Solve 3 sin x + 1 = 0 for 0° ≤ x ≤ 360°.
Step 1, isolate: 3 sin x = −1, so sin x = −1/3.
Step 2, basic angle: α = sin⁻¹(1/3) = 19.47° (to 2 d.p.).
Step 3, quadrants: sin x is negative, so x is in quadrant 3 or 4.
Step 4, build: x = 180° + 19.47° = 199.47° and x = 360° − 19.47° = 340.53°.
Answer: x = 199.5° and 340.5° (to 1 d.p.).
Check: sin 199.47° ≈ −0.3333 and sin 340.53° ≈ −0.3333. Both are inside the interval, and a sketch of y = sin x shows the line y = −1/3 crossing the curve twice between 0° and 360°.
The mistake to watch for
A common slip is to type sin⁻¹(−1/3) and stop at the calculator’s answer.
Mistaken answer: x = −19.5°
This is a genuine angle with sin equal to −1/3, but it lies outside the interval 0° ≤ x ≤ 360°, so it is not a valid answer. The other solution is also missing.
The correction is to use the calculator angle only as a clue. Take α = 19.47°, decide the quadrants from the negative sign, and build the two angles inside the interval. You can also add 360° to −19.47°, which gives 340.53° and agrees with the quadrant 4 result.
Check yourself
Give answers to 1 decimal place unless they are exact.
1. Solve cos x = 0.3 for 0° ≤ x ≤ 360°.
Show answer
Cosine is positive, so quadrants 1 and 4. α = cos⁻¹(0.3) = 72.54°. The angles are 72.54° and 360° − 72.54° = 287.46°.
x = 72.5° and 287.5°
2. Solve tan x = −2 for 0° ≤ x ≤ 360°.
Show answer
Tangent is negative, so quadrants 2 and 4. α = tan⁻¹(2) = 63.43°. Quadrant 2: 180° − 63.43° = 116.57°. Quadrant 4: 360° − 63.43° = 296.57°.
x = 116.6° and 296.6°
3. Solve 2 sin x = √3 for 0 ≤ x ≤ 2π, giving exact answers.
Show answer
sin x = √3/2, which is the exact value for π/3. Sine is positive, so quadrants 1 and 2: x = π/3 and x = π − π/3 = 2π/3.
x = π/3 and 2π/3
Where this leads next
Next, track a phase shift and period to see how the graph behind the equation changes. When the angle inside the function is not just x, find all roots after a variable substitution. The triangle and bearings reasoning board gives you a place to practise choosing a relationship before calculating, and the non-calculator working trainer keeps exact values ready.
If you can follow the method here but still miss solutions under exam pressure, our teachers can look at your working in online one-to-one Additional Mathematics tuition.