A curve C has equation y = 3sin 2x + 4cos 2x, -π ≤ x ≤ π - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 6

To express y in the form Rsin(2x + α), we first rewrite the given function:

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

We can use the formula:

$R = \sqrt{a^2 + b^2}$

where a = 3 and b = 4:

$R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Next, we find α using:

$\tan(α) = \frac{b}{a} = \frac{4}{3}$

Thus:

$α = \tan^{-1}(\frac{4}{3}) \approx 0.927$

This falls within the given constraints of 0 < α < π/2.

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

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

This leads to:

$4cos(2x) = -3sin(2x)$

Dividing both sides by cos(2x) (assuming cos(2x) ≠ 0):

$4 = -3tan(2x)$

This implies:

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

To solve for x, we find:

$2x = \tan^{-1}(-\frac{4}{3}) + nπ, n ∈ ℤ$

Thus:

$x = \frac{1}{2}\left(\tan^{-1}(-\frac{4}{3}) + nπ\right)$

Calculating the possible values:

1. For n = 0:

   • 2x ≈ -0.93 → x ≈ -0.465

2. For n = 1:

   • 2x ≈ 2.2 → x ≈ 1.1

Hence, the coordinates are approximately:

1. (-0.46, 0)
2. (1.11, 0)

Rounded to 2 decimal places.