Trigonometry: Question 4

Worked solution

Part (a): Solving $2\cos 2x - 1 = 0$

Rearrange to isolate the cosine term: $2\cos 2x - 1 = 0 \quad\Longrightarrow\quad \cos 2x = \frac12$

Extend the interval. Since $0^\circ \le x \le 360^\circ$, multiplying through by $2$ gives: $0^\circ \le 2x \le 720^\circ$

This is the crucial step: $2x$ must be treated as an angle ranging over a full $720^\circ$, not just $360^\circ$, so every solution of $\cos\theta = \tfrac12$ in that doubled range must be found before dividing by $2$.

Find all solutions for $2x$. The principal value of $\cos^{-1}\left(\tfrac12\right)$ is $60^\circ$. Cosine is also positive in the fourth quadrant, so within one $360^\circ$ cycle: $2x = 60^\circ \quad\text{or}\quad 2x = 360^\circ - 60^\circ = 300^\circ$

Since $2x$ can go up to $720^\circ$, add $360^\circ$ to each of these to capture the second cycle: $2x = 60^\circ + 360^\circ = 420^\circ \qquad 2x = 300^\circ + 360^\circ = 660^\circ$

So the complete set of values for $2x$ in $0^\circ \le 2x \le 720^\circ$ is: $2x = 60^\circ,\ 300^\circ,\ 420^\circ,\ 660^\circ$

Divide every value by $2$ to return to $x$: $x = 30^\circ,\ 150^\circ,\ 210^\circ,\ 330^\circ$

Check: at $x=30^\circ$, $2x=60^\circ$, $\cos 60^\circ = \tfrac12$ ✓. At $x=150^\circ$, $2x=300^\circ$, $\cos 300^\circ=\tfrac12$ ✓. At $x=210^\circ$, $2x=420^\circ\equiv60^\circ$, $\cos=\tfrac12$ ✓. At $x=330^\circ$, $2x=660^\circ\equiv300^\circ$, $\cos=\tfrac12$ ✓.

Part (b): Why there are 4 solutions, not 2

Solving $\cos\theta = \tfrac12$ on its own, for $0^\circ \le \theta \le 360^\circ$, gives exactly 2 solutions ($60^\circ$ and $300^\circ$).

Here, however, $\theta = 2x$, and because $x$ ranges over $0^\circ \le x \le 360^\circ$, the angle $2x$ ranges over $0^\circ \le 2x \le 720^\circ$, twice the usual $360^\circ$ span. Cosine repeats every $360^\circ$, so a doubled interval for $2x$ contains a second full cycle of cosine, which produces a second pair of solutions ($420^\circ$ and $660^\circ$) in addition to the first pair ($60^\circ$ and $300^\circ$). Dividing all 4 values of $2x$ by $2$ gives 4 distinct values of $x$, all lying within the original interval $0^\circ \le x \le 360^\circ$.

Final answers

• (a) $x = \boxed{30^\circ,\ 150^\circ,\ 210^\circ,\ 330^\circ}$
• (b) Doubling $x$ to get $2x$ doubles the interval searched (to $720^\circ$), which contains two full cycles of $\cos\theta = \tfrac12$ instead of one, giving 4 solutions instead of 2.