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Mathematics 15 Online
OpenStudy (yacoub1993):

Solve for the equation for the interval [0, 2pi) sin^2x-cos^2x=0

OpenStudy (anonymous):

i think i know the answer, hld on let me check

OpenStudy (yacoub1993):

ok

OpenStudy (anonymous):

heres the answer

OpenStudy (anonymous):

my choices are A)π4,π3 B)π4,π6 C)π4 D)π4,3π4,5π4,7π4

OpenStudy (yacoub1993):

same here

Directrix (directrix):

Try factoring: sin^2x-cos^2x=0 (sin x + cos x ) * (sin x - cos x ) = 0 Use the Zero Product Property

OpenStudy (yacoub1993):

is it A

Directrix (directrix):

(sin x + cos x ) * (sin x - cos x ) = 0 (sin x + cos x )= 0 OR (sin x - cos x ) = 0 sin x = - cos x OR sin x = cos x

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