How do you solve #sin 2 theta = cos theta#?

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Answer 1 · 2018-04-09T08:51:57

#theta=(2n+1)pi/2# or #theta=npi+-(-1)^npi/6#, where #n# is an integer

As #sin2theta=2sinthetacostheta#, we can write #sin2theta=costheta# as

#2sinthetacostheta=costheta#

or #2sinthetacostheta-costheta=0#

or #costheta(2sintheta-1)=0#

Hence either #costheta=0# i.e. #theta=(2n+1)pi/2#, where #n# is an integer

or #2sintheta-1=0# i.e. #sintheta=1/2=sin(pi/6)#

i.e. #theta=npi+-(-1)^npi/6#, where #n# is an integer