Consider $x \in(\pi; 2\pi)$ such that $\;3\cos\left(\dfrac x2\right)=-\sqrt2$
Determine the exact value of $\cos(x-\pi)+\cos\left(x-\dfrac\pi2\right)$.
I can't seem to find a way of solving this, Ii have tried using the angle summation formulas, etc.
By the duplication formula,
$$\cos x=2\left(-\frac{\sqrt2}{3}\right)^2-1=-\frac59=-\cos(x-\pi)$$ and $$\sin x=-2\frac{\sqrt2}{3}\sqrt{1-\left(-\frac{\sqrt2}{3}\right)^2}=-\frac{2\sqrt{14}}9=\cos(x-\frac\pi2).$$
(We take the positive sign for the sine because $x$ is known to exceed $\pi$.)