When a trig equation has something like 2x − 30° inside the function, replace that expression with a single letter, solve for the new letter on its new interval, then convert back to x. This appears in questions with sin 2x, cos 3x, tan(x + 45°) and in quadratics in sin x or cos x.
The method extends the basic angle and quadrant routine and uses the period ideas from tracking phase shift and period.
What changes when you substitute?
Write u = 2x − 30° (or whatever the bracket is). Two things follow.
- The interval for u is different. Transform the ends of the x interval with the same rule, so 0° ≤ x ≤ 360° becomes −30° ≤ u ≤ 690°.
- You solve for u, then convert. Find all u in the new interval using the basic angle method, then use x = (u + 30°)/2.
Widening the interval is how you keep the extra solutions. A multiplier of 2 doubles the number of cycles, so you expect about twice as many answers.
For a quadratic such as 2 sin²x − sin x − 1 = 0, the substitution is s = sin x. Solve the quadratic in s, then solve each sin x = s value on the interval. Discard any s outside −1 to 1.
Worked example
Solve sin(2x − 30°) = 0.5 for 0° ≤ x ≤ 360°.
Step 1, substitute: let u = 2x − 30°. Then x from 0° to 360° gives u from −30° to 690°.
Step 2, solve for u: sin u = 0.5 has basic angle 30°, in quadrants 1 and 2: u = 30° and 150°. Add 360° for the next cycle: u = 390° and 510°. The next ones, 750° and 870°, are above 690°, so stop. Going downwards, u = −210° and −330° are below −30°, so they are out.
Step 3, convert back: x = (u + 30°)/2.
- u = 30° gives x = 30°
- u = 150° gives x = 90°
- u = 390° gives x = 210°
- u = 510° gives x = 270°
Answer: x = 30°, 90°, 210°, 270°.
Check: sin(2 × 210° − 30°) = sin 390° = sin 30° = 0.5. Also sin(2 × 270° − 30°) = sin 510° = sin 150° = 0.5. Two full cycles of sine give four solutions, which matches.
The mistake to watch for
A common slip is to solve for u only between 0° and 360°.
Mistaken answer: x = 30° and 90°
The student used u = 30° and 150° only, because those are the sine solutions in the usual 0° to 360° range. The doubled angle means the interval for u reaches 690°, so the solutions at 390° and 510° were skipped.
The correction is to write the new interval for u before doing anything else, and list solutions until you pass its upper end. A short line such as “−30° ≤ u ≤ 690°” at the start of your working is a good habit.
Check yourself
1. Solve cos 3x = −0.5 for 0° ≤ x ≤ 180°.
Show answer
Let u = 3x, so 0° ≤ u ≤ 540°. cos u = −0.5: basic angle 60°, cosine negative in quadrants 2 and 3: u = 120° and 240°. Next cycle: 480°, and 600° is too large. Then x = u/3.
x = 40°, 80°, 160°
2. Solve tan(x + 45°) = 1 for 0° ≤ x ≤ 360°.
Show answer
Let u = x + 45°, so 45° ≤ u ≤ 405°. tan u = 1 at u = 45°, 225° and 405°. Subtract 45°.
x = 0°, 180°, 360°
3. Solve 2 sin²x − sin x − 1 = 0 for 0° ≤ x ≤ 360°.
Show answer
Let s = sin x: 2s² − s − 1 = 0, which factorises as (2s + 1)(s − 1) = 0, so s = −1/2 or s = 1. For sin x = 1, x = 90°. For sin x = −1/2, the basic angle is 30° in quadrants 3 and 4: x = 210° and 330°.
x = 90°, 210°, 330°
Where this leads next
Next, use a sketch to detect missing solutions and confirm your count before you submit. The quadratic structure explorer is useful for the quadratic-in-sin step, because the discriminant and roots of the quadratic in s decide how many values of sin x you must solve. The non-calculator working trainer keeps the exact values ready.
If you often reach the right method but lose solutions at the conversion step, our teachers can trace where in online one-to-one Additional Mathematics tuition.