6. (a) Express $3 \sin x + 2 \cos x$ in the form $R \sin(x + \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 5

To solve the equation $3 \sin x + 2 \cos x = 1$, we rewrite it using the expression in part (a):

$\sqrt{13} \sin(x + \alpha) = 1$

Dividing both sides by $\sqrt{13}$:

$\sin(x + \alpha) = \frac{1}{\sqrt{13}}$

Using the inverse sine function gives us:

$x + \alpha = \arcsin\left(\frac{1}{\sqrt{13}}\right) + 2n\pi$

and

$x + \alpha = \pi - \arcsin\left(\frac{1}{\sqrt{13}}\right) + 2n\pi$

from which we can solve for $x$:

For $0 < x < 2\pi$, let's calculate $\alpha$:

$$\alpha \approx \tan^{-1}\left(\frac{2}{3}\right) \approx 0.588.$

Therefore, substituting back in:

1. For the first case:

$x = \arcsin\left(\frac{1}{\sqrt{13}}\right) - 0.588$

2. For the second case:

$x = \pi - \arcsin\left(\frac{1}{\sqrt{13}}\right) - 0.588$

Calculating these values gives:

1. Approximating $\arcsin\left(\frac{1}{\sqrt{13}}\right) \approx 0.281.$ Thus: $x \approx 0.281 - 0.588 = -0.307 \text{ (not valid, discard)}$

2. Then for the second case:
   $x \approx \pi - 0.281 - 0.588 \approx 3.3.$

Continuing the checks within $2\pi$:
We also consider $x \approx 3.3 + 2\pi\approx 6.64$, which also needs validation.

Final valid answers for $0 < x < 2\pi$ rounded to 3 decimal places are thus: