Show that the equation \[2\sin^2\theta + 3\cos\theta = 3\] can be written as \[2\cos^2\theta - 3\cos\theta + 1 = 0.\] (2) Hence solve, for \(0 \le \theta \l...

Four marks on a trigonometric equation. The equation \(2\sin^2\theta + 3\cos\theta = 3\) mixes \(\sin\) and \(\cos\), which cannot be solved directly. The standard move is to use the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\) to eliminate one function in favour of the other. Here the \(\sin\) appears only as \(\sin^2\theta\), so replacing it with \(1 - \cos^2\theta\) leaves an equation in \(\cos\theta\) alone. Going the other way, replacing \(\cos\theta\) by \(\pm\sqrt{1 - \sin^2\theta}\), would introduce a square root and a sign ambiguity, which is why this direction is the right one.

(a) Show the equation can be written as \(2\cos^2\theta - 3\cos\theta + 1 = 0\) [2 marks]

The target is given, so all the credit is in the derivation. Substituting \(\sin^2\theta = 1 - \cos^2\theta\): \[2\bigl(1 - \cos^2\theta\bigr) + 3\cos\theta = 3 \quad \textbf{M1}\] Expanding and collecting everything on the left: \[2 - 2\cos^2\theta + 3\cos\theta - 3 = 0 \ \Rightarrow\ -2\cos^2\theta + 3\cos\theta - 1 = 0\] Multiplying through by \(-1\) to give a positive leading coefficient: \[2\cos^2\theta - 3\cos\theta + 1 = 0 \quad \textbf{A1}\] The M1 is for the correct use of the identity; the A1 is for reaching the printed form, including the sign change. Every intermediate line must be shown, since a candidate who writes down the given result unsupported earns nothing.

(b) Solve for \(0 \le \theta \lt 360^\circ\) [2 marks]

Treat the result as a quadratic in \(\cos\theta\). It factorises: \[\bigl(2\cos\theta - 1\bigr)\bigl(\cos\theta - 1\bigr) = 0 \ \Rightarrow\ \cos\theta = \tfrac{1}{2} \ \text{ or } \ \cos\theta = 1 \quad \textbf{M1}\] Now solve each. Cosine is positive in the first and fourth quadrants, so \(\cos\theta = \tfrac{1}{2}\) gives a principal value of \(60^\circ\) and a second solution at \(360^\circ - 60^\circ = 300^\circ\). The equation \(\cos\theta = 1\) has its only solution in the interval at \(\theta = 0^\circ\), the start of the range: \[\theta = 0^\circ, \ 60^\circ, \ 300^\circ \quad \textbf{A1}\] The M1 is for solving the quadratic to reach values of \(\cos\theta\), so it survives if only some of the angles are then found; the A1 requires all three angles and no extras. The interval notation is worth reading carefully: \(0^\circ\) is included because of the \(\le\), while \(360^\circ\) is excluded because of the strict inequality, so \(360^\circ\) must not be offered as a fourth solution.

Where the marks are

Only one of the four marks is a method mark, which makes this question unforgiving in part (a): the accuracy mark demands the exact printed rearrangement. In part (b), the danger is losing the A1 for an incomplete solution set rather than for a wrong one.

Common wrong turns on this question

• Using \(\sin^2\theta = 1 + \cos^2\theta\) or misremembering the identity, which produces the wrong quadratic.
• Dividing through by \(\cos\theta\) at some stage, which is invalid where \(\cos\theta = 0\) and destroys solutions.
• Missing \(\theta = 0^\circ\). Because \(\cos\theta = 1\) has a single solution in the range and it sits right on the boundary, it is the answer most often dropped. This is the omission the question is designed to catch.
• Offering \(\theta = 360^\circ\) as well. The range is \(\theta \lt 360^\circ\).
• Giving \(\theta = 240^\circ\) for \(\cos\theta = \tfrac{1}{2}\). That is the second solution for \(\cos\theta = -\tfrac{1}{2}\); for positive cosine the second solution is \(360^\circ - 60^\circ\).
• Working in radians. The interval is stated in degrees.

How to check

Substitute each angle into the original equation, not the rearranged one, so that an error in part (a) would be exposed. At \(\theta = 0^\circ\): \(2(0) + 3(1) = 3\), correct. At \(\theta = 60^\circ\): \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\), so \(2 \times \dfrac{3}{4} + 3 \times \dfrac{1}{2} = 1.5 + 1.5 = 3\), correct. At \(\theta = 300^\circ\): \(\sin 300^\circ = -\dfrac{\sqrt{3}}{2}\), and squaring removes the sign, so the value is again \(1.5 + 1.5 = 3\), correct. A final count check: a quadratic in \(\cos\theta\) with two distinct roots in \([-1, 1]\) usually yields up to four angles in a full revolution, but \(\cos\theta = 1\) is a boundary value contributing only one, which is why three solutions is the right total here.