Given that $f(x) = 7\cos x + \sin x$, where $R > 0$ and $0 < \alpha < 90^\circ$, (a) find the exact value of $R$ and the value of $\alpha$ to one decimal place - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 8

Rearranging the equation: $7\cos x + \sin x = 5$

We express it in terms of $R\cos (x - \alpha)$:

$\sqrt{50}\cos(x - 8.1^\circ) = 5$

Using $R = 5\sqrt{2}$, we have:

$\cos(x - 8.1^\circ) = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$

Thus, $x - 8.1^\circ = 45^\circ \implies x = 53.1^\circ$

Also, $x - 8.1^\circ = 315^\circ \implies x = 323.1^\circ$

The two solutions are:

• $x \approx 53.1^\circ$
• $x \approx 323.1^\circ$.

The equation $7\cos x + \sin x = k$ has only one solution when the line $y = k$ is tangent to the curve $y = 7\cos x + \sin x$.

The maximum value of $7\cos x + \sin x$ is: $R = 5\sqrt{2} \approx 7.07$

Hence, the only value of $k$ for which there is one solution must be: $k = \pm 5\sqrt{2}$

Thus, the required values are:

• $k = 5\sqrt{2}$
• $k = -5\sqrt{2}$.