HOW MANY SOLUTIONS DOES sin 2x = ½ HAVE BETWEEN 0° AND 360°?
There are four solutions. Doubling the angle doubles the interval the angle travels through, so 2x runs from 0° to 720° — two complete turns — and sin θ = ½ has two solutions in every turn.
How many solutions does sin 2x = ½ have in the interval 0° ≤ x < 360°?
Change the variable and change the interval with it. Everything goes wrong here at the point where the interval is left behind.
- let θ = 2x; as x runs over 0° ≤ x < 360°, θ runs over 0° ≤ θ < 720°
- sin θ = ½ gives θ = 30°, 150° in the first turn
- and θ = 390°, 510° in the second
- x = θ ÷ 2 = 15°, 75°, 195°, 255°
- that is 4 solutions
2 is what you get if you solve sin θ = ½ over 0° to 360° only and forget that θ is 2x, not x. Substituting without carrying the interval across is the error this question exists to catch, and it is the most common mistake in the whole of trigonometry at this level.
8 is what you get if you double the count as well as the interval The interval doubles once. Two turns give two pairs of solutions, not four.
For sin kx = c or cos kx = c with c strictly between −1 and 1, one full turn in x always gives exactly 2k solutions. Here k = 2, so the answer is 4 without solving anything. Tangent is the exception: tan kx = c gives k solutions per turn, because tan repeats every 180°, not every 360°.
THE GENERAL RULE
Whenever the argument of a trigonometric function is anything other than a bare x, transform the interval at the same moment you transform the variable, and write the new interval down before you solve. A question that gives you sin 2x rather than sin x is not testing trigonometry; it is testing whether you noticed.
Updated · written to the published ESAT content specification