How do you solve #2cos^2x=1# in the interval [0,360]?

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Answer 1 · 2016-11-22T08:09:54

The solutions are #S={45,135,225,315} #

Let's rewrite the equation

#2cos^2x-1=0#

#(sqrt2cosx+1)(sqrt2cosx-1)=0#

Therefore,

#(sqrt2cosx+1=0# , #=>#, #cosx=-1/sqrt2#

#x=135# and #x=225#

and #sqrt2cosx-1=0#, #=>#, #cosx=1/sqrt2#

#x=45# and #x=315#

The solutions, #S in[0,360]#

So, #S={45,135,225,315} #