Solving a Quadratic Trigonometric Equation

The examiner disguises a quadratic inside a trig equation. The plan is fixed: use a Pythagorean identity to get one trig ratio only, treat it as a quadratic, factorise, then solve each factor across the full range.

Solve $3\sin^2 x - 5\cos x - 1 = 0$ for $0^\circ \le x \le 360^\circ$. [6]

The working

Step 1, get a single ratio. The equation mixes $\sin^2 x$ and $\cos x$. Replace $\sin^2 x$ with $1 - \cos^2 x$: $3(1 - \cos^2 x) - 5\cos x - 1 = 0 \quad \text{(M1)}$ $3 - 3\cos^2 x - 5\cos x - 1 = 0 \;\Rightarrow\; -3\cos^2 x - 5\cos x + 2 = 0$

Multiply by $-1$ so the leading coefficient is positive: $3\cos^2 x + 5\cos x - 2 = 0 \quad \text{(A1)}$

Step 2, factorise as a quadratic in $\cos x$: $(3\cos x - 1)(\cos x + 2) = 0 \quad \text{(M1)}$

Step 3, solve each factor: $\cos x = \frac{1}{3} \qquad \text{or} \qquad \cos x = -2$

$\cos x = -2$ is impossible (cosine lies between $-1$ and $1$), so reject it, and say so. (A1 for $\cos x = \frac13$ and rejecting the other.)

Step 4, find every angle for $\cos x = \frac13$ in $0^\circ$ to $360^\circ$. The principal value is $x = 70.5^\circ$; cosine is also positive in the fourth quadrant, giving $360^\circ - 70.5^\circ$: $x = 70.5^\circ \quad \text{and} \quad x = 289.5^\circ \quad \text{(1 d.p.)} \quad \text{(A1, A1)}$

Where the marks are won and lost

• The identity swap must target the squared term so you end with one ratio. Replacing $\cos x$ instead of $\sin^2 x$ leaves a mess.
• Rejecting $\cos x = -2$ explicitly is expected. Silently ignoring it can cost a mark; carrying it forward certainly does.
• Both angles are required. Stopping at $70.5^\circ$ throws away the second accuracy mark, the fourth-quadrant solution is not optional.

Common mistakes

• Using $\sin^2 x = 1 + \cos^2 x$ (wrong sign).
• Cancelling a $\cos x$ or square-rooting instead of factorising, which loses solutions.
• Giving only the principal value and missing $289.5^\circ$.

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