Solving a Tangent Equation in a Range

Tangent has a $180^\circ$ period, so its solutions repeat every $180^\circ$ (you add $180^\circ$, not use the $\sin$/$\cos$ symmetries). Combined with a doubled argument, this question tests both ideas: widen the interval for the $2x$, and step by $180^\circ$ for the tan.

Solve $\tan(2x) = 1$ for $0^\circ \le x \le 180^\circ$. [4]

The working

Step 1, widen the interval. Let $u = 2x$; as $x$ runs $0^\circ$ to $180^\circ$, $u$ runs $0^\circ$ to $360^\circ$: $\tan u = 1, \qquad 0^\circ \le u \le 360^\circ \quad \text{(M1)}$

Step 2, solve $\tan u = 1$. The principal value is $45^\circ$; tangent repeats every $180^\circ$, so add $180^\circ$: $u = 45^\circ, \ 225^\circ \quad \text{(M1, A1)}$

Step 3, divide by 2 (since $u = 2x$): $x = 22.5^\circ, \ 112.5^\circ \quad \text{(A1)}$

Where the marks are won and lost

• Tan’s period is $180^\circ$, so the second solution is $45^\circ + 180^\circ = 225^\circ$, not a $360^\circ$-based reflection. Using sine/cosine symmetry here gives wrong angles.
• Widen the interval to $0^\circ$–$360^\circ$ for $u = 2x$ first, then divide the solutions by $2$. Solving over the $x$-range directly misses solutions.
• Two solutions are expected (two $180^\circ$-spaced values in a $360^\circ$ window).

Common mistakes

• Treating tan like sin/cos and using $180 - 45$ (giving $135^\circ$) for the second solution.
• Forgetting to widen the interval for the doubled argument.
• Dividing the interval instead of the solutions.

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