f(x) = 5 ext{cos}x + 12 ext{sin}x Given that f(x) = R ext{cos}(x - ext{α}), where R > 0 and 0 < α < \frac{π}{2} - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 5

Substituting R and α into the equation:

$\cos(x - \alpha) = \frac{6}{R}$

We have:

$\cos(x - 1.176) = \frac{6}{13}$

Using the arccos function:

$x - \alpha = \arccos{\left(\frac{6}{13}\right)}$

Calculating:

$x - 1.176 \approx 1.091$

Thus:

$x \approx 1.091 + 1.176 \implies x \approx 2.267 \text{ or }\quad \forall x = 2\pi - (x - \alpha) \implies x \approx 2 ext{π} - 1.091 + 1.176$. So, the values of x that satisfy the equation in the interval are approximately 2.267 and (2π - 1.091 + 1.176).

The maximum value of the function can be derived as:

$R_{max} = 13$

Thus, the maximum value of 5cosx + 12sinx is 13.

The maximum occurs when:

$\cos(x - \alpha) = 1 \implies x - \alpha = 0 \implies x = \alpha$

And since we have found \alpha \approx 1.176,

The smallest positive value of x for which this maximum occurs is:

$x \approx 1.176$.