2026-08-31T11:35:12.387260 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 3 shows part of the curve \(C\) with equat...

This question tests the second Pythagorean identity, the one connecting the cotangent and the cosecant, and then asks you to read the number of solutions off the printed graph rather than count them by trial. The graph is not decoration: part (b) explicitly requires it as the justification.

(a) Showing that the equation may be written as \(2\operatorname{cosec}^{2}x + 5\operatorname{cosec} x - 12 = 0\) [2 marks]

The equation \(2\cot^{2}x + 5\operatorname{cosec} x = 10\) contains two different functions, so it cannot be solved as it stands. Dividing \(\sin^{2}x + \cos^{2}x \equiv 1\) through by \(\sin^{2}x\) gives \(1 + \cot^{2}x \equiv \operatorname{cosec}^{2}x\), which rearranges to

\[\cot^{2}x \equiv \operatorname{cosec}^{2}x - 1.\]

That is the identity to use, because it converts the squared term into the same function as the linear term. Substituting:

\[2\left(\operatorname{cosec}^{2}x - 1\right) + 5\operatorname{cosec} x = 10 \qquad \textbf{M1}\] \[2\operatorname{cosec}^{2}x - 2 + 5\operatorname{cosec} x = 10 \quad \Longrightarrow \quad 2\operatorname{cosec}^{2}x + 5\operatorname{cosec} x - 12 = 0 \qquad \textbf{A1}\]

Since the result is printed, the M1 is for quoting and using the correct identity and the A1 for reaching the given form exactly, constant term included. The characteristic error is using the wrong Pythagorean relation, for instance \(\cot^{2}x = 1 - \operatorname{cosec}^{2}x\), which flips a sign and gives \(-2\operatorname{cosec}^{2}x + 5\operatorname{cosec} x - 8 = 0\). The other is arithmetic: the constant is \(-2 - 10 = -12\), not \(-8\).

(b) Solving the equation and explaining why there are exactly four solutions [4 marks]

Treat \(\operatorname{cosec} x\) as a single unknown; the expression is a quadratic in it and factorises:

\[\left(2\operatorname{cosec} x - 3\right)\left(\operatorname{cosec} x + 4\right) = 0 \qquad \textbf{M1}\]

The M1 is for a correct method of solution, so the quadratic formula earns it equally, and it is a follow-through from part (a). The two roots are \(\operatorname{cosec} x = \dfrac{3}{2}\) and \(\operatorname{cosec} x = -4\). Now invert, since \(\operatorname{cosec} x = \dfrac{1}{\sin x}\) and so \(\sin x = \dfrac{1}{\operatorname{cosec} x}\):

\[\sin x = \frac{2}{3} \qquad \text{or} \qquad \sin x = -\frac{1}{4} \qquad \textbf{A1}\]

Both are admissible because each has modulus at most \(1\). This inversion is where marks are most often lost: writing \(\sin x = \tfrac{3}{2}\), which is impossible, or \(\sin x = -4\), reverses the reciprocal. A useful guard is that the cosecant never takes a value strictly between \(-1\) and \(1\), so a root of the quadratic in that band would have to be rejected, whereas here both roots are outside it and both survive.

For \(\sin x = \dfrac{2}{3}\) the sine is positive, so \(x\) is in the first or second quadrant. The principal value is \(41.81^{\circ}\) and its partner is \(180^{\circ} - 41.81^{\circ}\):

\[x = 41.8^{\circ},\ 138.2^{\circ} \qquad \textbf{A1}\]

For \(\sin x = -\dfrac{1}{4}\) the sine is negative, so \(x\) is in the third or fourth quadrant. With an acute reference angle of \(14.48^{\circ}\), the solutions are \(180^{\circ} + 14.48^{\circ}\) and \(360^{\circ} - 14.48^{\circ}\):

\[x = 194.5^{\circ},\ 345.5^{\circ} \qquad \textbf{A1}\]

All four values are given to one decimal place as instructed, and all four lie strictly inside \(0^{\circ} \lt x \lt 360^{\circ}\). Note that \(x = 180^{\circ}\) is excluded in the question because \(\operatorname{cosec} x\) is undefined there, which is why the interval is written the way it is.

Reading the count from Figure 3. The explanation is part of what these four marks pay for, so a bare list of values is not a complete answer to the part. Solving \(\operatorname{cosec} x = \tfrac{3}{2}\) and \(\operatorname{cosec} x = -4\) amounts to intersecting the curve \(C\) with the two horizontal lines \(y = \tfrac{3}{2}\) and \(y = -4\). Figure 3 shows that the branch on \(0^{\circ} \lt x \lt 180^{\circ}\) lies entirely at or above \(y = 1\), dipping to a minimum of \(1\), so a line at height \(\tfrac{3}{2}\) lies above that minimum and cuts the branch twice, while a line at \(-4\) misses it altogether. The branch on \(180^{\circ} \lt x \lt 360^{\circ}\) lies entirely at or below \(y = -1\), rising to a maximum of \(-1\), so the line \(y = -4\) cuts it twice and the line \(y = \tfrac{3}{2}\) misses it. Two intersections plus two intersections gives exactly four solutions, which is precisely the list found algebraically.

That is also the check on the algebra. If the working had produced a value of \(\operatorname{cosec} x\) between \(-1\) and \(1\), the graph would show no intersection at all at that height, revealing the slip at once.

Examination takeaway. Learn \(1 + \cot^{2}x \equiv \operatorname{cosec}^{2}x\) alongside the more familiar Pythagorean identity, and take the reciprocal carefully when returning from the cosecant to the sine. When a printed graph is offered, use it to justify the number of solutions: the horizontal line \(y = k\) meets the curve as often as the equation has roots.