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Mathematics 13 Online
OpenStudy (moongazer):

sec^2 x = 1 what is the value of x?

OpenStudy (ash2326):

sec x = 1/ cos x sec^2x=1 so cos^2 x=1 or \[ cos x= \pm 1\]

OpenStudy (ash2326):

cos x = +1 or -1 so x =0 or pi so the solution is x= n*pi n =0,1,2 and so on

OpenStudy (moongazer):

thanks

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