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Mathematics 16 Online
OpenStudy (anonymous):

Solve the equation on the interval [0, 2π). sin^2x - cos^2x = 0 A. π/4 B. π/4 ; π/6 C. π/4 ; π/3 D. π/4 ; 3π/4 ; 5π/4 ; 7π/4

OpenStudy (anonymous):

D should work since \(\sin^2(x)=\cos^2(x)\) means \(\sin(x)=\pm\cos(x)\)

OpenStudy (anonymous):

Thank you!

OpenStudy (anonymous):

at all those numbers you get \(\pm\frac{\sqrt{2}}{2}\) for sine and cosine

OpenStudy (anonymous):

yw

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