The curve C has the equation $$ ext{cos }2x + ext{cos }3y = 1, $$ $$ -rac{ ext{pi}}{4} ext{ }< ext{x} < rac{ ext{pi}}{4} ext{ and } 0 < ext{y} < rac{ ext{pi}}{6} - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 7

Substituting ( x = \frac{\text{pi}}{6} ) into the curve equation:

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

This simplifies to:

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

which gives:

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

Thus:

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

Therefore:

$$3y = \frac{\text{pi}}{3} ext{ or } 3y = \frac{5\text{pi}}{3}.$$

This leads to:

$$y = \frac{\text{pi}}{9} ext{ or } y = \frac{5\text{pi}}{9}.$$

Since ( y < \frac{\text{pi}}{6} ), we select ( y = \frac{\text{pi}}{9}. )

From part (a), we have:

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

Calculating these values:

$$\sin(\frac{\text{pi}}{3}) = \frac{\sqrt{3}}{2}, \quad \sin(\frac{\text{pi}}{3}) = \frac{\sqrt{3}}{2}.$$

Thus:

$$\frac{dy}{dx} = \frac{2 \cdot \frac{\sqrt{3}}{2}}{3 \cdot \frac{\sqrt{3}}{2}} = \frac{2}{3}.$$

The point P has coordinates ( \left( \frac{\text{pi}}{6}, \frac{\text{pi}}{9} \right) ). Using the point-slope form of the equation of the line:

$$(y - y_1) = m(x - x_1),$$

where ( m = \frac{2}{3} ):

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

Clearing the fractions and rearranging gives:

$$6y - 2\text{pi} = 4x - \text{pi},$$

which can be expressed as:

$$4x - 6y + 2\text{pi} = 0.$$

Thus, the final form is:

4x - 6y + 2\text{pi} = 0 \text{, where } a = 4, b = -6, c = 2.\