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

At the point (P), where (x = \frac{\pi}{6}):

1. Substitute (x = \frac{\pi}{6}) into the original equation:
   [\text{cos} \left(2 \cdot \frac{\pi}{6}\right) + \text{cos} 3y = 1]
   [\text{cos} \left(\frac{\pi}{3}\right) + \text{cos} 3y = 1]
   [\frac{1}{2} + \text{cos} 3y = 1]
2. Rearranging gives:
   [\text{cos} 3y = 1 - \frac{1}{2} = \frac{1}{2}]
3. Therefore, (3y) can be calculated as follows:
   [3y = \frac{\pi}{3} + 2k\pi \quad (k \in \mathbb{Z})]
4. This leads to:
   [y = \frac{\pi}{9} + \frac{2k\pi}{3}]

1. We already know that at (x = \frac{\pi}{6}), (y = \frac{\pi}{9}).

2. Evaluate (\frac{dy}{dx}) at this point:
   [\frac{dy}{dx} = \frac{2\sin(\frac{\pi}{3})}{3\sin(\frac{\pi}{3})} = \frac{2}{3}]

3. The slope (m = \frac{2}{3}).

4. Using point-slope form (y - y_1 = m(x - x_1)):
   [y - \frac{\pi}{9} = \frac{2}{3}\left(x - \frac{\pi}{6}\right)]

5. Rearranging gives:
   [2x - 3y + \frac{\pi}{2} - \frac{\pi}{3} = 0]

6. To express it in the required form:
   [6x - 9y - \frac{\pi}{2} = 0]

   Multiplying through appropriately will provide integer coefficients. The equation is then stated as:
   [6x + 9y - 2\pi = 0]

   This gives integers a, b, and c.