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

To find the value of y at the point P where $x = \frac{\pi}{6}$, we substitute this value into the curve's equation:

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

This simplifies to:

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

resulting in:

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

Solving for $\cos 3y$ gives:

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

Thus, we have:

$3y = \frac{\pi}{3} + 2k\pi \text{ or } 3y = -\frac{\pi}{3} + 2k\pi,$

where k is any integer. However, since $0 \leq y < \frac{\pi}{6}$, we only consider:

$3y = \frac{\pi}{3} \Rightarrow y = \frac{\pi}{9}.$

To find the equation of the tangent at P, we first evaluate $\frac{dy}{dx}$ using the value of $x = \frac{\pi}{6}$:

$\frac{dy}{dx} = \frac{2\sin\left(2\cdot\frac{\pi}{6}\right)}{3\sin\left(3\cdot\frac{\pi}{9}\right)} = \frac{2\sin\left(\frac{\pi}{3}\right)}{3\sin\left(\frac{\pi}{3}\right)} = \frac{2}{3}.$

The slope of the tangent line at point P (where $x = \frac{\pi}{6}$, $y = \frac{\pi}{9}$) can be expressed as:

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

where $(x_1, y_1) = \left(\frac{\pi}{6}, \frac{\pi}{9}\right)$ and $m = \frac{2}{3}$.

Substituting these values gives:

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

Rearranging this yields:

$y = \frac{2}{3}x + b.$

By calculating b, we can express the tangent equation in the necessary form. After rearranging, we find the form becomes:

$6x + 9y - 2\pi = 0,$

which corresponds to integers a, b, and c.