Functions, f and g, are given by $f(x) = 3 + ext{cos} \, x$ and $g(x) = 2x \, , x \, ext{in} \, ext{R}$ - Scottish Highers Maths - Question 6 - 2018

Next, we find the expression for $g(f(x))$ by substituting $f(x)$ into $g(x)$:

$g(f(x)) = g(3 + ext{cos}(x)) = 2(3 + ext{cos}(x)) = 6 + 2 ext{cos}(x).$

Thus, the final expression is:

$g(f(x)) = 6 + 2 ext{cos}(x).$

We set the two expressions equal to each other:

$3 + ext{cos}(2x) = 6 + 2 ext{cos}(x).$

Rearranging gives:

$ext{cos}(2x) - 2 ext{cos}(x) = 3.$ Substituting the identity $ext{cos}(2x) = 2 ext{cos}^2(x) - 1$ leads to:

$2 ext{cos}^2(x) - 1 - 2 ext{cos}(x) = 3.$ This simplifies to:

$2 ext{cos}^2(x) - 2 ext{cos}(x) - 4 = 0.$ Factoring gives:

$2( ext{cos}^2(x) - ext{cos}(x) - 2) = 0.$ Using the quadratic formula:

$ext{cos}(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2}.$
This yields solutions for $x$. Ensure that the solutions are within the domain $0 \leq x < 2\pi$.