Question

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

Answer 1

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

Explanation:

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`

Answer 2

`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!

Answer 3

`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`