Solve #cos^2theta=1/2#, if #theta# lies in the interval #[0,2pi)#?

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Answer 1 · 2017-12-15T17:11:04

#theta={pi/4,(3pi)/4,(5pi)/4,(7pi)/4}#

We can write #cos^2theta=1/2# as

#cos^2theta=(1/sqrt2)^2#

or #cos^2theta-(1/sqrt2)^2=0#

or #(costheta-1/sqrt2)(costheta+1/sqrt2)=0#

i.e. #costheta=+-1/sqrt2=+-cos(pi/4)#

Hence #theta=npi+-pi/4# where #n# is an integer and in the interval #[0,2pi)#

#theta={pi/4,(3pi)/4,(5pi)/4,(7pi)/4}#

Answer 2 · 2017-12-16T00:15:43

#color(blue)(pi/4)# and #color(blue)((7pi)/4)#

#cos^2theta=1/2#

#costheta=sqrt(1/2)=1/sqrt(2)=sqrt(2)/2#

For: #[0,2pi)#

#arccos(sqrt(2)/2)=color(blue)(pi/4)# and #2pi-pi/4=color(blue)((7pi)/4)#