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 5 - 2015 - Paper 3

Using the previously derived form:

$4cos2θ + 2sin2θ = 1$

We substitute to find:

$\frac{1}{R} = \frac{1}{2\sqrt{5}}$,

And simplify:

$cos(2θ - 26.57°) = \, \frac{1}{√20}$.

Then,

$cos(2θ - 26.57°) = \frac{1}{\sqrt{20}}$ implies:

$2θ - 26.57° = \pm \arccos\left(\frac{1}{\sqrt{20}}\right)$.

Calculating the arc cosine:

$θ ≈ 1.81°,\, (first\, solution)$

For the second solution:

$2θ - 26.57° = 180° + \arccos\left(\frac{1}{\sqrt{20}}\right)$,

Thus:

$θ = \frac{180° + \arccos\left(\frac{1}{\sqrt{20}}\right) + 26.57°}{2}$.

After calculation, this gives us another value of:

$θ ≈ 25.35°$ (second solution).

From our earlier calculations:

Given that g(θ) has no solutions, We deduce the conditions for k as:

1. From the formula for R, the maximum value must be: $k ≤ R = 2\sqrt{5}$, which simplifies approximately to:

   $k ≤ 4.47$.

2. Since for no solutions $k$ should also stay above the minimum possible, we find: $k ≥ -√20$

Therefore, the range for k is:

$-√20 < k < 2√5$,

or numerically approximately:

$-4.47 < k < 4.47$.