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

Question

Solve the equation on the interval $$0 \leq \theta < 2 \pi$$
-$\cos ( 2 \theta ) = \sqrt { 2 } - \cos ( 2 \theta )$
A) $\frac { \pi } { 8 } , \frac { 7 \pi } { 8 } , \frac { 9 \pi } { 8 } , \frac { 15 \pi } { 8 }$

B) $\frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 }$

C) $0 , \frac { 2 \pi } { 3 } , \pi , \frac { 4 \pi } { 3 }$

D) No solution

Quizplus · Accepted Answer

First, solve the equation $\cos(2\theta) = \sqrt{2} - \cos(2\theta)$ by adding $\cos(2\theta)$ to both sides to get $2\cos(2\theta) = \sqrt{2}$, which simplifies to $\cos(2\theta) = \frac{\sqrt{2}}{2}$. The solutions to $\cos(2\theta) = \frac{\sqrt{2}}{2}$ on the interval $0 \leq \theta < 2\pi$ are $\theta = \frac{\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{15\pi}{8}$ because $\cos(2\theta) = \frac{\sqrt{2}}{2}$ corresponds to the angles where $2\theta$ equals $45^\circ$ (or $\frac{\pi}{4}$), $135^\circ$ (or $\frac{3\pi}{4}$), $225^\circ$ (or $\frac{5\pi}{4}$), and $315^\circ$ (or $\frac{7\pi}{4}$) respectively. Solving for $\theta$ gives the angles listed in option A.

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

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