Solve #2(sinx+cosx)+2sin2x+1=0# ?

Using de Moivre's identity

#e^(ix) = cosx+isinx#

#2((e^(ix)-e^(-ix))/(2i)+(e^(ix)+e^(-ix))/2)+2((e^(2ix)-e^(-2ix))/(2i))+1=0# or

#((e^(ix)-e^(-ix))/(i)+e^(ix)+e^(-ix))+((e^(2ix)-e^(-2ix))/(i))+1=0# or

#e^(ix)-e^(-ix)+i(e^(ix)+e^(-ix))+e^(2ix)-e^(-2ix)+i=0# or

#(e^(ix)-e^(-ix))(1+e^(ix)+e^(-ix))+i(e^(ix)+e^(-ix)+1) = 0# or

#(e^(ix)+e^(-ix)+1)(e^(ix)-e^(-ix)+i)=0 # or

#(2cosx+1)(2sin x+1)=0#

so the solutions are

#cosx = -1/2# and #sin x = -1/2# or

#(x = 2/3pi+2kpi) uu (x = 7/6pi+2kpi)# for #k = 0,pm1,pm2,cdots#

#2sinx + 2cosx + 2sin2x + 1 = 0#

#2sinx + 2cosx + 2(2sinxcosx) + 1= 0#

#2sinx + 2cosx + 4sinxcosx + 1 = 0#

#(2sinx + 2cosx)^2 = (-1 - 4sinxcosx)^2#

#4sin^2x + 4cos^2x + 8sinxcosx = 1 + 8sinxcosx + 16sin^2xcos^2x#

#4(sin^2x + cos^2x) + 8sinxcosx - 1 - 8sinxcosx - 16sin^2xcos^2x = 0#

#4 - 1 - 16sin^2xcos^2x = 0#

#3 = 16sin^2x(1 - sin^2x)#

#3 = 16sin^2x - 16sin^4x#

#0 = -16sin^4x + 16sin^2x - 3#

#0 = -16sin^4x + 4sin^2x + 12sin^2x - 3#

#0 = -4sin^2x(4sin^2x - 1) + 3(4sin^2x - 1)#

#0= (-4sin^2x + 3)(4sin^2x- 1)#

#sinx= +- sqrt(3)/2" AND "sinx = +-1/2#

#x = 60˚, 120˚, 240˚, 300˚, 30˚, 150˚, 210˚, 330˚#

However, instantly, checking in the original equation, you will notice many of the solutions are extraneous. The actual solutions are as follows:

# x = 210˚, x = 330˚,x = 120˚#

Hopefully this helps!

#2(sinx+cosx)+2sin2x+1=0#

#=>2sinx+2cosx+4sinxcosx+1=0#

#=>2sinx+4sinxcosx+2cosx+1=0#

#=>2sinx(1+2cosx)+(1+2cosx)=0#

#=>(1+2cosx)(2sinx+1)=0#

when
#(1+2cosx)=0#
#=>cosx=-1/2=cos((2pi)/3)#
#=>x=2npi+-(2pi)/3" "where" "ninZZ#

Again when

#2sinx+1=0#
#=>sinx=-1/2=sin(-(pi)/6)#
#=>x=npi-(-1)^npi/6" "where" "ninZZ#