A curve C has equation y = 3sin 2x + 4cos 2x, -\pi < x < \pi - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 6

To express y in the required form, we can use the technique of combining sine and cosine. We start from:

$y = 3\sin(2x) + 4\cos(2x).$

We can express it as:

$y = R\sin(2x + \alpha)$

where

$R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5.$

To find \alpha, we use:

$\tan(\alpha) = \frac{4}{3} \, \Rightarrow \, \alpha = \tan^{-1}\left(\frac{4}{3}\right) \approx 0.927.$

Thus, \alpha = 0.927 (to 3 significant figures).

To find the intersection points with the x-axis, we set y = 0:

$0 = 3\sin(2x) + 4\cos(2x).$

This can be rewritten as:

$\sin(2x) = -\frac{4}{3}\cos(2x).$

Dividing both sides by cos(2x):

$\tan(2x) = -\frac{4}{3}.$

Now we find 2x:

$2x = \tan^{-1}\left(-\frac{4}{3}\right) + n\pi, \text{ for } n \in \mathbb{Z}.$

Calculating, we find the solutions for x:

$x \approx -2.03, -0.46, 1.11, 2.68.$

Therefore, the coordinates are:

$(-2.03, 0), (-0.46, 0), (1.11, 0), (2.68, 0)$

All answers rounded to 2 decimal places.