A curve C has equation $$y = 3 ext{sin}2x + 4 ext{cos}2x, \, -rac{ ext{π}}{2} < x < ext{π}.$$ The point A(0, 4) lies on C - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 6

To express y in the form Rsin(2x + α), we combine the sine and cosine terms:

$y = 3\text{sin}2x + 4\text{cos}2x.$

Using the R and α definitions:

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

To find α, we use:

$\tan{α} = \frac{4}{3}.$

Thus,

$α = \text{arctan}\left(\frac{4}{3}\right) \approx 0.927.$

The intersection with the x-axis occurs where y = 0:

$0 = 3\text{sin}2x + 4\text{cos}2x.$

We can rewrite this as:

$\text{sin}(2x + α) = 0,$

where we previously found that α is associated with the coefficients. This gives us:

$2x + α = n\text{π}, \, n \in \mathbb{Z}.$

Solving for x:

$x = \frac{n\text{π} - α}{2}.$

Substituting values for n = -2, -1, 0, 1 and calculating:

• For n = -2: $x = \frac{-2\text{π} - 0.927}{2} \approx -3.50$
• For n = -1: $x = \frac{-\text{π} - 0.927}{2} \approx -1.57$
• For n = 0: $x = \frac{0 - 0.927}{2} \approx -0.46$
• For n = 1: $x = \frac{\text{π} - 0.927}{2} \approx 1.11$
• For n = 2: $x = \frac{2\text{π} - 0.927}{2} \approx 2.68$

Thus the coordinates of intersection with the x-axis are approximately:

$(-3.50, 0), (-1.57, 0), (-0.46, 0), (1.11, 0), (2.68, 0).$