6. (a) Express $3 \, ext{sin} \, x + 2 \, ext{cos} \, x$ in the form $R \, ext{sin}(x + heta)$ where $R > 0$ and $0 < heta < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 5

The expression $R \, \text{sin}(x + \alpha)$ has a maximum value of $R$ since the range of sine function is from -1 to 1. Here, the maximum value of $3 \, ext{sin} \, x + 2 \, ext{cos} \, x$ is $\sqrt{13}$. Thus, the greatest value of $(3 \, \text{sin} \, x + 2 \, \text{cos} \, x)^4$ is:

$\left(\sqrt{13}\right)^4 = 13^2 = 169.$

To solve the equation $3 \, \text{sin} \, x + 2 \, \text{cos} \, x = 1$, we can rewrite it as: $\sqrt{13} \, \text{sin}(x + 0.588) = 1,$ which simplifies to: $\text{sin}(x + 0.588) = \frac{1}{\sqrt{13}} \approx 0.277.$

The solutions for $\text{sin} \theta = k$ are given by: $\theta = \sin^{-1}(k) + 2k\pi \text{ and } \theta = \pi - \sin^{-1}(k) + 2k\pi.$

Thus, calculating for our case:

1. First solution: $x + 0.588 = \sin^{-1}(0.277) \approx 0.281 + 2k\pi,\ k=0\implies x \approx 0.281 - 0.588 = -0.307.$ (Disregard since not in range.)
2. Second solution: $x + 0.588 = \pi - \sin^{-1}(0.277) \approx \pi - 0.281 \approx 2.860\implies x \approx 2.860 - 0.588 = 2.272.$
3. For the second cycle: $x + 0.588 = 2\pi - \sin^{-1}(0.277) \approx 6.10 \implies x \approx 6.10 - 0.588 = 5.513.$

Therefore, the values of $x$ that satisfy the equation are approximately:

• $x \approx 2.272$
• $x \approx 5.513$ (to three decimal places).