Solve the equation cos x = sin 2x, in the interval 0 <= x <= pi. Express 5 cos x - 12 sin x in the form Rcos(x + a), where R > 0 and 0 < a < 90°, giving the value of a to one decimal place. Hence or otherwise, find the minimum and maximum values of 10 / (5 cos x - 12 sin x + 15).

Case 1: $\cos x = 0$
In the interval $0 \le x \le \pi$, the solution is $x = \frac{\pi}{2}$.

Case 2: $1 - 2\sin x = 0$

In the interval $0 \le x \le \pi$, the solutions are $x = \frac{\pi}{6}$ and $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

The solutions are $\boxed{x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}}$.

To one decimal place, $\alpha = 67.4^\circ$. This value is within $0^\circ < \alpha < 90^\circ$.
So, $5\cos x - 12\sin x = \boxed{13\cos(x+67.4^\circ)}$.

The expression becomes $\frac{10}{|13\cos(x+67.4^\circ)| + 15}$.

We know that $-1 \le \cos(x+67.4^\circ) \le 1$.
Therefore, $0 \le |\cos(x+67.4^\circ)| \le 1$.
Multiplying by 13, we get $0 \le |13\cos(x+67.4^\circ)| \le 13$.

Minimum value of denominator: $0 + 15 = 15$.
Maximum value of denominator: $13 + 15 = 28$.

The fraction is maximized when the denominator is minimized.
Maximum value $= \frac{10}{15} = \frac{2}{3}$.
The fraction is minimized when the denominator is maximized.
Minimum value $= \frac{10}{28} = \frac{5}{14}$.

The minimum value is $\boxed{\frac{5}{14}}$ and the maximum value is $\boxed{\frac{2}{3}}$.

3 de faites, encore 2 aujourd'hui. Tu progresses.