Solve the equation on the interval [0, 2π).
-$$\sin ^ { 2 } x - \cos ^ { 2 } x = 0$$
A) $\frac { \pi } { 4 } , \frac { 3 \pi } { 4 } , \frac { 5 \pi } { 4 } , \frac { 7 \pi } { 4 }$
B) $\frac { \pi } { 4 } , \frac { \pi } { 6 }$
C) $\frac { \pi } { 4 } , \frac { \pi } { 3 }$
D) $\frac { \pi } { 4 }$

Question

Quizplus · Accepted Answer

The equation $\sin^2x - \cos^2x = 0$ simplifies to $\sin^2x = \cos^2x$, which implies $\tan^2x = 1$. The solutions to $\tan x = \pm 1$ in the interval [0, 2π) are $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.

Solve the Equation on the Interval [0, 2π) sin⁡2x−cos⁡2x=0\sin ^ { 2 } x - \cos ^ { 2 } x = 0sin2x−cos2x=0

Solve the Equation on the Interval [0, 2π) $\sin ^ { 2 } x - \cos ^ { 2 } x = 0$