Trigonometry: Question 3

Worked solution

Part (a): Rewriting the equation

Start from the identity connecting $\sin^2 x$ and $\cos^2 x$: $\sin^2 x + \cos^2 x \equiv 1 \quad\Longrightarrow\quad \sin^2 x \equiv 1 - \cos^2 x$

Substitute this into the left-hand side of $2\sin^2 x = 1 + \cos x$: $2(1 - \cos^2 x) = 1 + \cos x$

Expand the bracket: $2 - 2\cos^2 x = 1 + \cos x$

Collect every term on one side (subtract $1+\cos x$ from both sides): $2 - 2\cos^2 x - 1 - \cos x = 0$

$1 - 2\cos^2 x - \cos x = 0$

Multiply through by $-1$ so that the $\cos^2 x$ term is positive: $2\cos^2 x + \cos x - 1 = 0$

as required.

Part (b): Solving the quadratic in $\cos x$

Let $c = \cos x$, so the equation becomes: $2c^2 + c - 1 = 0$

Factorise: $(2c - 1)(c + 1) = 0$

(Check: $(2c-1)(c+1) = 2c^2 + 2c - c - 1 = 2c^2 + c - 1$ ✓.)

So either: $2c - 1 = 0 \Rightarrow c = \cos x = \frac12 \qquad\text{or}\qquad c + 1 = 0 \Rightarrow c = \cos x = -1$

Case 1: $\cos x = \dfrac12$

The principal value is $x = 60^\circ$. Since cosine is also positive in the fourth quadrant, the second solution in $0^\circ \le x \le 360^\circ$ is: $x = 360^\circ - 60^\circ = 300^\circ$

Case 2: $\cos x = -1$

This has exactly one solution in a full $360^\circ$ interval: $x = 180^\circ$

Collecting all solutions in $0^\circ \le x \le 360^\circ$: $x = 60^\circ,\ 180^\circ,\ 300^\circ$

Check (substituting back into the original equation): at $x=60^\circ$, $2\sin^2 60^\circ = 2\left(\tfrac{\sqrt3}{2}\right)^2 = \tfrac32$ and $1+\cos 60^\circ = 1+\tfrac12=\tfrac32$ ✓. At $x=180^\circ$, both sides equal $0$ ✓. At $x=300^\circ$, both sides equal $\tfrac32$ ✓.

Final answers

• (a) $2\cos^2 x + \cos x - 1 = 0$, as shown above.
• (b) $x = \boxed{60^\circ,\ 180^\circ,\ 300^\circ}$