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Mathematics 16 Online
OpenStudy (anonymous):

Solve sin^2 x = cos^2 x for 0 degrees less than or equal to x less than 360

OpenStudy (anonymous):

Sin^2 x= cos^2 x 0 <= x < 360 Sub (1 - sin^2) for cos^2 sin^2 = 1 - sin^2 2sin^2 = 1 sin^2 = 1/2 ---- sin = -sqrt(2)/2, +sqrt(2)/2 x = 45 degrees, 135 degrees, 225 degrees, 315 degrees is this correct?

OpenStudy (anonymous):

not sure about it, because: cos^2x-sin^2x=cos2x=0 cos t=0 for t=90 and t=270 so x=45 and x=135

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

@onegirl, looks ok.. and if you're unsure, check your answers by plugging your answers back into the original equation....

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