g(θ) = 4cos 2θ + 2sin 2θ Given that g(θ) = R cos(2θ - α), where R > 0 and 0 < α < 90°, (a) find the exact value of R and the value of α to 2 decimal places - Edexcel - A-Level Maths Pure - Question 4 - 2015 - Paper 3

Starting with the equation:

4cos2θ + 2sin2θ = 1.\

We can substitute the earlier results to find:

$2cos2θ + sin2θ = \frac{1}{2}.$

Rearranging gives:

$2cos2θ + \frac{1}{2} = 2$

Thus,

• We can employ:

2cos2θ = 1 - 2sin2θ\

From the cosine identity, we can express:

$cos(2θ) = \frac{1}{D}.$

Next, we find the possible values for θ by solving:

• Finding primary solutions:
  • Using the cosine values and solving for θ gives:
    • First solution: θ ≈ 0.90°
    • Second solution: θ ≈ -0.44°.

To determine the range of possible values of k where the equation g(θ) = k has no solutions, we realize that for k to have no solutions, it must contradict the maximum and minimum values of g(θ).

• The maximum value of g(θ):
  • From the earlier calculation, we determined that the maximum value can be derived from the values: k must be within the range of:
  • minimum k: k < -√20
  • maximum k: k > √20.

Thus, the allowable range for k is:

$k < -√20 \text{ or } k > √20.$