How do you find all solutions in the internal [0,2pi) for 5cos 2theta=0 ?

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Answer 1 · 2018-04-01T13:32:36

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

As #5cos2theta=0#, we have #cos2theta=0#

or #2cos^2theta-1=0#

i.e. #cos^2theta=1/2#

or #costheta=+-1/sqrt2=+-cos(pi/4)#

and therefore general solution is #theta=npi+-pi/4#

and in the interval #[0,2pi)#, we have

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