How do you solve #Sin (4 theta)= cos(2theta)#?

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Answer 1 · 2016-05-02T13:01:49

#theta=(2n+1)pi/4# or

#theta=npi/2+(-1)^npi/12#

#sin(4theta)=cos(2theta)#

or #2sin(2theta)cos(2theta)-cos(2theta)=0#

or #cos(2theta){2sin(2theta)-1}=0#

i.e. either #cos(2theta)=0# or #2sin(2theta)-1=0#

If #cos(2theta)=0#, #2theta=(2n+1)pi/2# or #theta=(2n+1)pi/4#

and if #2sin(2theta)-1=0#, we have #sin(2theta)=1/2#

and #2theta=npi+(-1)^npi/6# or #theta=npi/2+(-1)^npi/12#