How do I solve #cos2x+3cosx=-2#?

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Answer 1 · 2017-02-28T20:48:54

#x_1=npi, n=2k+-1, k in ZZ#

#x_2=n/3pi,mod(n,2)=0#

#cos2x+3cosx=-2#

Use the double angle formula for cosine to expand #cos2x# and rewrite the equation in standard form

#2cos^2x-1+3cosx+2=0#

#2cos^2x+3cosx+1=0#

Let #cosx=y#

#y=(-3+-sqrt(9-4(2)(1)))/(2(2))=(-3+-1)/4#

#y=-1# or #y=-1/2#

#cosx=-1#

#x=npi, n=2k+-1, k in ZZ#

#cosx=-1/2#

#x_2=n/3pi,mod(n,2)=0#