Solve the equation #2cos2x+1/2sin2x=sin^2x#?

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Answer 1 · 2017-10-12T05:32:37

#x=npi+pi/4# or #x=npi+tan^(-1)(-2/3)#, where #n# is an integer

#2cos2x+1/2sin2x=sin^2x# can be written as

#2(1-2sin^2x)+sinxcosx=sin^2x#

or #5sin^2x-sinxcosx-2=0# and dividing each term by #cos^2x#

#5tan^2x-tanx-2(1+tan^2x)=0#

or #3tan^2x-tanx-2=0#

i.e. #(3tanx+2)(tanx-1)=0#

i.e. either #tanx-1=0=>tanx=1=tan(pi/4)# i.e. #x=npi+pi/4#

or #3tanx+2=0=>tanx=-2/3# i.e. #x=npi+tan^(-1)(-2/3)#, where #n# is an integer