Solve, for \(0 \leqslant x < 360^\circ\), the equation \(2\sin(x + 40^\circ) = 1\). (4) Solve, for \(0 \leqslant \theta < 180^\circ\), the equation \(\tan 2...

Both parts are trigonometric equations in which the argument is not simply \(x\) or \(\theta\) but a transformed angle: \(x + 40^\circ\) in the first and \(2\theta\) in the second. The reliable technique for these is always the same. Transform the given interval into an interval for the whole argument, find every solution of the simple equation inside that transformed interval, and only then undo the transformation. Working the other way round, by solving for the base angle first and adjusting afterwards, is what causes solutions to go missing.

(a) Solve \(2\sin(x + 40^\circ) = 1\) for \(0 \le x \lt 360^\circ\) [4]

Divide by \(2\) to isolate the sine:

\[\sin(x + 40^\circ) = \frac{1}{2} \qquad \textbf{[M1]}\]

The [M1] is for reaching \(\sin(\text{argument}) = \tfrac{1}{2}\). Now transform the interval. Adding \(40^\circ\) to each part of \(0 \le x \lt 360^\circ\) gives

\[40^\circ \le x + 40^\circ \lt 400^\circ \qquad \textbf{[M1]}\]

The second [M1] is for this adjusted interval, and it is the mark that makes the question work. The interval for the argument extends past \(360^\circ\), so a solution beyond one full revolution has to be included.

Now solve \(\sin\phi = \tfrac{1}{2}\) for \(40^\circ \le \phi \lt 400^\circ\). The principal value is \(\phi = 30^\circ\), which is below \(40^\circ\) and so is not in range, but its partner \(180^\circ - 30^\circ = 150^\circ\) is. Adding a full revolution to the rejected value gives \(30^\circ + 360^\circ = 390^\circ\), which is in range:

\[x + 40^\circ = 150^\circ \quad\text{or}\quad x + 40^\circ = 390^\circ \qquad \textbf{[A1]}\] \[x = 110^\circ \quad\text{or}\quad x = 350^\circ \qquad \textbf{[A1]}\]

The first [A1] is for both values of the argument, the second for both values of \(x\). The error the question is built to catch is stopping at \(x = 110^\circ\): the extended interval carries a second solution, and it is worth exactly one mark. Note also that \(30^\circ\) itself is not a solution, since \(x = -10^\circ\) lies outside the given interval. Working in degrees is correct throughout, because the interval is stated in degrees.

(b) Solve \(\tan 2\theta = -\sqrt{3}\) for \(0 \le \theta \lt 180^\circ\) [4]

Transform the interval first. Doubling each part of \(0 \le \theta \lt 180^\circ\) gives

\[0 \le 2\theta \lt 360^\circ \qquad \textbf{[M1]}\]

The [M1] is for this doubled interval. Doubling the interval is essential: a full revolution for \(2\theta\) corresponds to only half a revolution for \(\theta\), so all the solutions must be gathered before halving.

The related acute angle comes from \(\tan 60^\circ = \sqrt{3}\) [M1]. Since the required tangent is negative, and the tangent is negative in the second and fourth quadrants, the solutions are \(180^\circ - 60^\circ\) and \(360^\circ - 60^\circ\):

\[2\theta = 120^\circ \quad\text{or}\quad 2\theta = 300^\circ \qquad \textbf{[A1]}\] \[\theta = 60^\circ \quad\text{or}\quad \theta = 150^\circ \qquad \textbf{[A1]}\]

Two independent accuracy marks, one for the pair of values of \(2\theta\) and one for the pair of values of \(\theta\). The tangent function repeats every \(180^\circ\), not every \(360^\circ\), which is why \(120^\circ\) and \(300^\circ\) differ by exactly \(180^\circ\); that periodicity is the quickest way to generate the second solution once the first is known.

The trap here is the sign. A calculator returns \(-60^\circ\) for \(\arctan\left(-\sqrt{3}\right)\), which is outside the interval, and a candidate who simply halves it offers \(\theta = -30^\circ\) and scores no accuracy marks. Take the acute angle \(60^\circ\) from the positive value, then place the solutions in the correct quadrants according to the sign. A second trap is halving the interval instead of doubling it, which restricts \(2\theta\) to \(0 \le 2\theta \lt 90^\circ\) and finds nothing at all.

Check: substitute the answers back. For part (a), \(2\sin(110^\circ + 40^\circ) = 2\sin 150^\circ = 2 \times 0.5 = 1\), and \(2\sin(350^\circ + 40^\circ) = 2\sin 390^\circ = 2\sin 30^\circ = 1\). For part (b), \(\tan(2 \times 60^\circ) = \tan 120^\circ = -1.732 = -\sqrt{3}\), and \(\tan(2 \times 150^\circ) = \tan 300^\circ = -1.732\). All four values satisfy their equations and all four lie inside the stated intervals. A useful counting rule: over a full revolution a sine or cosine equation normally has two solutions, and a doubled argument normally doubles that count, so finding two solutions in each part here is exactly what should be expected.