Solve the equation #cos6x-cos4x=-sinx#?

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Answer 1 · 2018-01-22T06:58:17

#x=npi# or #x=(npi)/5+(-1)^npi/30#, where #n# is an integer.

As #cosA-cosB=-2sin((A+B)/2)sin((A-B)/2)#

we can write #cos6x-cos4x=-sinx# as

#-2sin5xsinx=-sinx#

or #2sin5xsinx-sinx=0#

or #sinx(2sin5x-1)=0#

therefore either #sinx=0# or #2sin5x-1=0#

if #sinx=0#, #x=npi#

and if #2sin5x-1=0#, #sin5x=1/2=sin(pi/6)#

i.e. #5x=npi+(-1)^npi/6# an #x=(npi)/5+(-1)^npi/30#

Here #n# is an integer.