Question

If cosxcos(pi/7)+sinxsin(pi/7)=-sqrt(2)/2, then which of these could x equal?

1. 7pi/4 + pi/7 + 2npi
2. 5pi/4 + pi/7 + 2npi
3. pi/4 + pi/7 + 2npi
4. 3pi/4 + pi/7 + 2npi

Answer 1

\[\begin{align*}\cos x\cos\frac{\pi}{7}+\sin x\sin\frac{\pi}{7}&=-\frac{\sqrt2}{2}\\
\cos\left(x-\frac{\pi}{7}\right)&=-\frac{1}{\sqrt2}\\
x-\frac{\pi}{7}&=\cos^{-1}\left(-\frac{1}{\sqrt2}\right)
\end{align*}\]
In the unit circle, the RHS gives you the angles \(\dfrac{3\pi}{4}\) and \(\dfrac{5\pi}{4}\). The general solution will thus be either angle plus an additional rotation of \(2\pi\) radians in either direction, i.e. \(\dfrac{3\pi}{4}+2n\pi\) and \(\dfrac{5\pi}{4}+2n\pi\) for integers \(n\).

So,
\[x=\frac{3\pi}{4}+\frac{\pi}{7}+2n\pi~~\text{or}~~\frac{5\pi}{4}+\frac{\pi}{7}+2n\pi\]