Solve the equation for solutions in the interval [0, 2 $$[ 0,2 \pi ) .$$
-$\sqrt { 2 } \cos 2 x = 1$
A) $\left\{ 0 , \frac { 2 \pi } { 3 } , \pi , \frac { 4 \pi } { 3 } \right\}$
B) $\left\{ \frac { \pi } { 8 } , \frac { 9 \pi } { 8 } , \frac { 7 \pi } { 8 } , \frac { 15 \pi } { 8 } \right\}$
C) $\varnothing$
D) $\left\{ \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 } \right\}$

Question

Quizplus · Accepted Answer

First, divide both sides by $\sqrt{2}$ to get $\cos 2x = \frac{1}{\sqrt{2}}$, which simplifies to $\cos 2x = \cos \frac{\pi}{4}$. The solutions to $\cos 2x = \cos \frac{\pi}{4}$ in the interval $[0, 2\pi)$ are $x = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}, \frac{15\pi}{8}$. However, since we need to solve for $2x$, the correct solutions within the given interval are $x = \frac{\pi}{8}, \frac{9\pi}{8}, \frac{7\pi}{8}, \frac{15\pi}{8}$.

Solve the Equation for Solutions in the Interval [0, 2 [0,2π).[ 0,2 \pi ) .[0,2π).

Solve the Equation for Solutions in the Interval [0, 2 $[ 0,2 \pi ) .$