Solve the equation #cos2x=-2cos^2x-3cosx#?

3 Answers
Shwetank Mauria
Aug 1, 2017

#x=(2n+1)xx180^@#
or #x=2nxx180^@+-75.5^@#, where #n# is an integer.

Explanation:

As #cos2x=2cos^2x-1#,

#cos2x=-2cos^2x-3cosx# is

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

or #4cos^2x+3cosx-1=0#

or #4cos^2x+4cosx-cosx-1=0#

i.e. #4cosx(cosx+1)-1(cosx+1)=0#

or #(4cosx-1)(cosx-1)=0#

i.e. #cosx=1# or #cosx=-1/4=cos75.5^@#

Hence, #x=(2n+1)xx180^@# or #x=2nxx180^@+-75.5^@#, where #n# is an integer.

P dilip_k
Aug 1, 2017

#cos2x=2cos^2x-3cosx#

#=>2cos^2x-1=-2cos^2x-3cosx#

#=>4cos^2x+3cosx-1=0#

#=>4cos^2x+4cosx-cosx-1=0#

#=>4cosx(cosx+1)-1(cosx+1)=0#

#=>(4cosx-1)(cosx+1)=0#

when

#4cosx-1=0#

#=>4cosx=1#

#=>cosx=1/4 =cos(0.42pi)#

#=>x=2npipm0.42pi" where "n in ZZ#

when

#4cosx-1=0#

#=>cosx+1=0#

#=>cosx=-1#

#=>x=(2n+1)pi" where "n in ZZ#

Cesareo R.
Aug 1, 2017

#{(x = pmpi + 2kpi),(x = pm arctan sqrt(15) + 2kpi):} #

for #k = 0, pm1,pm2,pm3, cdots#

Explanation:

Reducing to the fundamental period with #cos(2x)=1-2cos^2x#

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

#4cos^2x+3cosx-1=0# or

#{(cosx=-1->x = pmpi + 2kpi),(cosx=1/4->x = pm arctan sqrt(15) + 2kpi):} #

for #k = 0, pm1,pm2,pm3, cdots#

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