How do you find the x intercepts of #y=sec^4((pix)/8)-4#?

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Answer 1 · 2017-01-31T13:25:49

#x=4n-2#, #n=1,2,3,#...

#y=sec^4((pix)/8)-4#

at #x# intercepts, #y=0#
#sec^4((pix)/8)-4=0#

#(sec^2((pix)/8)-2)(sec^2((pi x)/8)+2)=0#

#(sec^2((pi x)/8)+2)=0#, #sec^2((pi x)/8)=-2#, cannot solve.

#sec^2((pix)/8)-2=0#, #sec((pix)/8)=+-sqrt2#

#(pix)/8=pi/4, (3pi)/4, (5pi)/4,...#

#x=2,6,10,...#
#x=4n-2#, #n=1,2,3,#...