Solve the equation on the interval $$0 \leq \theta

Question

Solve the equation on the interval $$0 \leq \theta < 2 \pi$$
-$\sqrt { 2 } \cos ( 2 \theta ) = 1$
A) $\frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 }$
B) $0 , \frac { 2 \pi } { 3 } , \pi , \frac { 4 \pi } { 3 }$
C) $\frac { \pi } { 8 } , \frac { 7 \pi } { 8 } , \frac { 9 \pi } { 8 } , \frac { 15 \pi } { 8 }$
D) No solution

Quizplus · Accepted Answer

First, solve the equation $\sqrt{2} \cos(2\theta) = 1$ by dividing both sides by $\sqrt{2}$, giving $\cos(2\theta) = \frac{1}{\sqrt{2}}$. This simplifies to $\cos(2\theta) = \cos(\frac{\pi}{4})$, because $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. The general solution for $\cos(2\theta) = \cos(\frac{\pi}{4})$ is $2\theta = 2n\pi \pm \frac{\pi}{4}$, where $n$ is an integer. Solving for $\theta$ gives $\theta = n\pi \pm \frac{\pi}{8}$. Considering the interval $0 \leq \theta < 2\pi$, the solutions are $\frac{\pi}{8}$, $\frac{7\pi}{8}$, $\frac{9\pi}{8}$, and $\frac{15\pi}{8}$.

Solve the Equation on the Interval 0≤θ<2π0 \leq \theta < 2 \pi0≤θ<2π

Solve the Equation on the Interval $0 \leq \theta < 2 \pi$