Solving a Trig Equation with a Multiple Angle

When the argument is $2x$ (or $3x$, or $x + 30^\circ$), the golden rule is to transform the interval to match the argument, solve there, then transform back. A doubled argument means twice the range, and therefore twice as many solutions, missing them is the defining error of this question type.

Solve $\sin(2x) = 0.5$ for $0^\circ \le x \le 360^\circ$. [5]

The working

Step 1, widen the interval. Let $u = 2x$. As $x$ runs $0^\circ$ to $360^\circ$, $u$ runs $0^\circ$ to $720^\circ$: $\sin u = 0.5, \qquad 0^\circ \le u \le 720^\circ \quad \text{(M1)}$

Step 2, find every solution of $\sin u = 0.5$ in $[0^\circ, 720^\circ]$. The principal value is $30^\circ$; sine is also positive in the second quadrant ($150^\circ$); then add $360^\circ$ for the second cycle: $u = 30^\circ,\ 150^\circ,\ 390^\circ,\ 510^\circ \quad \text{(M1, A1)}$

Step 3, divide each by 2 (since $u = 2x$): $x = 15^\circ,\ 75^\circ,\ 195^\circ,\ 255^\circ \quad \text{(A1, A1)}$

Where the marks are won and lost

• Widen the interval first. Solving $\sin u = 0.5$ over only $0^\circ$–$360^\circ$ finds $u = 30^\circ, 150^\circ$ and gives just $x = 15^\circ, 75^\circ$, missing half the answers. The argument’s range is the key.
• Find all solutions for $u$ before dividing, it’s easier to spot the full set in the widened interval than to hunt for extras afterward.
• Four solutions are expected here (two per $360^\circ$ cycle, two cycles). If you have two, you forgot to widen.

Common mistakes

• Solving over the $x$-range instead of the $2x$-range.
• Dividing the interval by 2 instead of multiplying (it should widen, not shrink).
• Missing the second-quadrant solution within each cycle.

Full method: Solving Trig Equations notes. Topic home: Trigonometry pillar.