Use Identities to Solve Trigonometric Equations
Solve the equation on the interval [0, 2π).
-$\sin ^ { 2 } 2 x = 1$
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) $\frac { \pi } { 8 } , \frac { 9 \pi } { 8 }$
D) no solution

Question

Quizplus · Accepted Answer

$\sin^2 2x = 1$ implies that $\sin 2x = \pm 1$. This occurs at angles where the sine function reaches its maximum and minimum values, which are at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ within one cycle. Since $2x$ is the argument, we divide these angles by 2 to find the solutions for $x$, resulting in $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ within the interval [0, 2π).

Use Identities to Solve Trigonometric Equations Solve the Equation on the Interval

Use Identities to Solve Trigonometric Equations
Solve the Equation on the Interval