The curve C has the equation $$ ext{cos }2x + ext{cos }3y = 1, $$ where $$ -rac{ ext{pi}}{4} leq x leq rac{ ext{pi}}{4}, 0 leq y leq rac{ ext{pi}}{6} - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 7

At the point P where $x = \frac{\text{pi}}{6},$ we substitute:

$$\text{cos }\left(2 \times \frac{\text{pi}}{6}\right) + \text{cos }3y = 1.$$

Calculating:

$$\text{cos}\left(\frac{\text{pi}}{3}\right) + \text{cos }3y = 1\Rightarrow \frac{1}{2} + \text{cos }3y = 1.$$

This simplifies to:

$$\text{cos }3y = \frac{1}{2}.$$

Solving for 3y yields:

$$3y = \frac{\text{pi}}{3} + 2n\pi,$$

where n is an integer. Since $0 \tleq y \tleq \frac{\text{pi}}{6},$ we take:

y = \frac{\text{pi}}{9}.$

At $x = \frac{\text{pi}}{6}$, we have previously found:

$$\frac{dy}{dx} = \frac{2\sin(2\times \frac{\text{pi}}{6})}{3\sin(3\times \frac{\text{pi}}{9})} = \frac{2\sin(\frac{\text{pi}}{3})}{3\sin(\frac{\text{pi}}{3})} = \frac{2}{3}.$$

The point P at $\left(\frac{\text{pi}}{6}, \frac{\text{pi}}{9}\right)$ allows us to use point-slope form:

$$\text{y} - \frac{\text{pi}}{9} = \frac{2}{3}\left(x - \frac{\text{pi}}{6}\right).$$

Rearranging gives:

$$2x - 3y + \frac{\text{pi}}{9} - \frac{\text{pi}}{9} = 0\Rightarrow 2x - 3y - \frac{\text{pi}}{9} = 0.$$

Multiplying through by 9 for integer coefficients yields:

18x - 27y - \text{pi} = 0,$$ where coefficients a, b, and c are 18, -27, and -1 respectively.