01 · Two turns
cos 2x = −1/2, 0° ≤ x < 360°
Hint
Use 0° ≤ 2x < 720°.
Worked solution
2x = 120°, 240°, 480°, 600°
x = 60°, 120°, 240°, 300°
Understand · explore · practise
Multiple-angle and shifted-angle equations
Solve equations such as sin(2x − 30°) = 1/2 by mapping the whole interval. Handle shifts, negative multipliers and unknown frequencies without losing solutions.
Before you startBasic trigonometric equations and linear inequalities
01 / Name the whole angle
Solve for the argument before solving for x.
In sin(2x − 30°), the sine receives the complete argument 2x − 30°. Set u = 2x − 30°. You now have an ordinary sine equation in u, but you must also transform the given x-interval.
For f(ax + b) = k:
1. Set u = ax + b.
2. Map the entire x-interval to u.
3. Find every u solution in that interval.
4. Convert back using x = (u − b)/a.
Here a ≠ 0. Keep track of open and closed endpoints.
The model shows both intervals and pairs every retained argument with its x-value. Change the equation to compare a multiple, a shift and a reversed interval.
Map the interval and every solutionChoose and compare
Let u = 2x − 30°. x in [0°, 360°) maps to u in [-30°, 690°). Pairs (u → x): 30° → 30°; 150° → 90°; 390° → 210°; 510° → 270°.
Watch an argument interval expand before mapping back
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / A multiple of x
Two turns of the argument can fit into one turn of x.
Solve sin 2x = 1/2 for 0° ≤ x < 360°. Set u = 2x, so 0° ≤ u < 720°. The sine curve completes two turns over this interval.
u = 30°, 150°, 390°, 510°
x = u/2 = 15°, 75°, 195°, 255°
If you stop at u = 30°, 150°, you lose two valid answers. Do not divide the first angle by 2 and then apply “180° minus x”: the symmetry belongs to the argument u.
03 / A shifted angle
Translate the interval along with the angle.
Solve cos(x + 40°) = 1/2 for 0° ≤ x < 360°. With u = x + 40°, the interval is 40° ≤ u < 400°.
cos u = 1/2
u = 60°, 300°
x = u − 40° = 20°, 260°
Starting with an automatic 0° to 360° range for u happens to retain the same candidates in this example, but it is not the method. A shift can move a needed solution into a neighbouring turn.
A shifted boundary solution
For sin(x + 30°) = 1/2 on 0° ≤ x ≤ 360°, the argument interval is [30°, 390°]. Keep u = 30°, 150°, 390°, giving x = 0°, 120°, 360°. The final 390° candidate would be missed by restricting u to the first turn.
04 / Multiply and shift
Map both endpoints before generating the families.
Solve sin(2x − 30°) = 1/2 for 0° ≤ x < 360°.
The full argument intervalWorked example
u = 2x − 30°
−30° ≤ u < 690°
Apply the same affine expression to the two endpoints.
u = 30°, 150°, 390°, 510°
Both sine families are needed; every value lies in the mapped interval.
x = (u + 30°)/2
x = 30°, 90°, 210°, 270°
Convert every candidate, then check the original x-interval.
You can instead write x = 30° + 180°n or x = 90° + 180°n and select the permitted integers. Both methods describe the same answer set.
05 / A negative multiplier
Reverse the inequality when the interval flips.
Solve cos(60° − 2x) = 1/2 for −30° ≤ x ≤ 150°. At x = −30°, u = 120°; at x = 150°, u = −240°. Put the resulting endpoints in increasing order:
−240° ≤ u ≤ 120°
u = −60°, 60°
x = (60° − u)/2 = 60°, 0°
Sorted in increasing order, the answers are x = 0°, 60°. If an endpoint is open, its exclusion follows it when the order reverses. For a = 0, there is no inverse mapping: evaluate the constant f(b), then either every permitted x works or none does.
06 / Tangent and complements
Keep the function’s own period.
Solve tan(3x + 15°) = 1 for 0° ≤ x < 180°. The argument interval is 15° ≤ u < 555°. Tangent repeats every 180°.
u = 45°, 225°, 405°
x = (u − 15°)/3 = 10°, 70°, 130°
A complement identity can change the appearance of an equation. For example, sin(90° − 2x) = 1/2 is cos 2x = 1/2. On 0° ≤ x < 180°, the answers are x = 30°, 150°.
If the equation contains a sine/cosine ratio of the same argument, first check whether its cosine can be zero, then convert to tangent and use its 180° period.
07 / An unknown frequency
One observed solution rarely determines the multiplier.
Suppose k > 0 and x = 20° is a solution of sin(kx) = 1/2. This tells us sin(20k°) = 1/2. It does not tell us that 20k must be the principal angle.
20k = 30 + 360n or 150 + 360n
k = 3/2 + 18n or 15/2 + 18n
Here n = 0,1,2,… gives all positive k in both families. There are infinitely many choices. The smallest is k = 3/2. An extra condition such as a specified period, range for k or first positive solution may determine a unique value.
08 / Your turn
Keep the argument and x on separate lines.
All intervals and answers are in degrees.
02 · A shift
sin(x − 20°) = 1/2, 0° ≤ x < 360°
Hint
−20° ≤ u < 340°.
Worked solution
u = 30°, 150°
x = 50°, 170°
03 · Combined
sin(3x + 30°) = 1/2, 0° ≤ x < 180°
Hint
30° ≤ u < 570°.
Worked solution
u = 30°, 150°, 390°, 510°
x = 0°, 40°, 120°, 160°
04 · A reversed interval
cos(90° − 2x) = 0, 0° ≤ x ≤ 180°
Hint
−270° ≤ u ≤ 90°.
Worked solution
u = −270°, −90°, 90°
x = 180°, 90°, 0°
In increasing order: 0°, 90°, 180°.
05 · Tangent
tan(2x − 30°) = −1, −90° ≤ x < 90°
Hint
−210° ≤ u < 150°, with u = −45° + 180°n.
Worked solution
u = −45°, 135°
x = −7.5°, 82.5°
06 · An open endpoint
sin(x + 30°) = 1/2, 0° < x < 360°
Hint
Both mapped endpoints are excluded.
Worked solution
30° < u < 390°
u = 150° ⇒ x = 120°
07 · A slow argument
cos(x/2) = 0, −360° ≤ x ≤ 360°
Hint
−180° ≤ u ≤ 180°.
Worked solution
u = −90°, 90°
x = −180°, 180°
08 · Unknown k
k > 0 and x = 30° solves cos(kx) = 1/2. Find the smallest possible k and explain why it is not unique.
Hint
30k = ±60 + 360n.
Worked solution
k = 2 + 12n or −2 + 12n
Keep n ≥ 0 in the first family and n ≥ 1 in the second. The smallest positive value is 2; infinitely many larger values also work.
09 · Find the mistake
A student solving sin 3x = 1/2 on 0° ≤ x < 180° writes x = 10° or 170°. Correct the answer.
Hint
Apply sine symmetry to 3x, not to x.
Worked solution
0° ≤ 3x < 540°
3x = 30°, 150°, 390°, 510°
x = 10°, 50°, 130°, 170°
09 / Recap
Transform the interval as well as the expression.
- Name the complete argument u.
- Map both endpoints and preserve whether each is included.
- A negative multiplier reverses the interval order.
- Solve across the full u-range before converting back.
- Tangent repeats every 180°; sine and cosine every 360°.
- One observed solution can leave infinitely many possible frequencies.
Section 1 of 9 · Name the whole angle