Solve the equation for exact solutions over the interval [0, 2 $$[ 0,2 \pi )$$
-$$\sin ^ { 2 } x - \cos ^ { 2 } x = 0$$
A) $\left\{ \frac { \pi } { 4 } , \frac { \pi } { 3 } \right\}$
B) $\left\{ \frac { \pi } { 4 } \right\}$
C) $\left\{ \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 } \right\}$
D) $\left\{ \frac { \pi } { 4 } , \frac { \pi } { 6 } \right\}$

Question

Quizplus · Accepted Answer

The equation $\sin^2x - \cos^2x = 0$ can be rewritten using the Pythagorean identity $\sin^2x + \cos^2x = 1$ as $2\sin^2x - 1 = 0$ or $2\cos^2x - 1 = 0$, leading to $\sin^2x = \cos^2x$. This equation is satisfied when $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ within the interval $[0, 2\pi)$.

Solve the Equation for Exact Solutions Over the Interval [0 [0,2π)[ 0,2 \pi )[0,2π)

Solve the Equation for Exact Solutions Over the Interval [0 $[ 0,2 \pi )$