3. (a) Express $5 \, ext{cos} \, x - 3 \, ext{sin} \, x$ in the form $R \text{cos}(x + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2} \pi$.\n\n(b) Hence, or otherwise, solve the equation\n$5 \, ext{cos} \, x - 3 \, ext{sin} \, x = 4$\nfor $0 \leq x < 2\pi$, giving your answers to 2 decimal places. - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 2

To convert the expression $5 \, \text{cos} \, x - 3 \, \text{sin} \, x$ into the form $R \text{cos}(x + \alpha)$, we start by equating coefficients based on the cosine and sine components.\n\n1. Find $R$: We have the coefficients as $A = 5$ and $B = -3$. The magnitude $R$ can be calculated using the formula:
$R = \sqrt{A^2 + B^2} = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.831 \text{ (to three decimal places)}$\n\n2. Find $\alpha$: To find $\alpha$, we use the tangent ratio:
$\tan(\alpha) = \frac{B}{A} = \frac{-3}{5} \Rightarrow \alpha = \arctan\left(\frac{-3}{5}\right) \approx -0.5404 \text{ radians}$\n Since $\alpha$ must be positive, we can consider the angle in appropriate quadrants if necessary. Hence, we will write our expression as: $5 \text{cos} x - 3 \text{sin} x = \sqrt{34} \text{cos}(x + 0.5404).$