2cos^2x-1=0,x=5pi/4?

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Answer 1 · 2018-05-03T20:35:47

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if your question was

#2cos^2(x)-1=0#

The solution:

hit the two side by #1/2#

#cos^2(x)-1/2=0#

#cos^2(x)=1/2#

#cos(x)=+-1/sqrt2#

#x=45# when #cosx=1/sqrt2#

#x=315# when #cosx=1/sqrt2#

#x=135# when #cosx=-1/sqrt2#

#x=225# when #cosx=-1/sqrt2#

or you mean

#2cos^2(x-1)=#

#cos^2(x-1)=#

#x=(5pi/4)=225#

#cos^2(225-1)=#

#cos^2(224)=0.34127#