Solve $2 (sin x + cos x) + 2 sin 2 x + 1 = 0$ $2(sinx+cosx)+2sin2x+1=0$ ?

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Solve $2 (sin x + cos x) + 2 sin 2 x + 1 = 0$ $2(sinx+cosx)+2sin2x+1=0$ ?

3 Answers

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Answer 1 · 2016-09-23T08:27:49

$(x = \frac{2}{3} π + 2 k π) \cup (x = \frac{7}{6} π + 2 k π)$ $(x = 2/3pi+2kpi) uu (x = 7/6pi+2kpi)$ for $k = 0, \pm 1, \pm 2, \dots$ $k = 0,pm1,pm2,cdots$

Explanation:

Using de Moivre's identity

$e^{i x} = cos x + i sin x$ $e^(ix) = cosx+isinx$

$2 (\frac{e^{i x} - e^{- i x}}{2 i} + \frac{e^{i x} + e^{- i x}}{2}) + 2 (\frac{e^{2 i x} - e^{- 2 i x}}{2 i}) + 1 = 0$ $2((e^(ix)-e^(-ix))/(2i)+(e^(ix)+e^(-ix))/2)+2((e^(2ix)-e^(-2ix))/(2i))+1=0$ or

$(\frac{e^{i x} - e^{- i x}}{i} + e^{i x} + e^{- i x}) + (\frac{e^{2 i x} - e^{- 2 i x}}{i}) + 1 = 0$ $((e^(ix)-e^(-ix))/(i)+e^(ix)+e^(-ix))+((e^(2ix)-e^(-2ix))/(i))+1=0$ or

$e^{i x} - e^{- i x} + i (e^{i x} + e^{- i x}) + e^{2 i x} - e^{- 2 i x} + i = 0$ $e^(ix)-e^(-ix)+i(e^(ix)+e^(-ix))+e^(2ix)-e^(-2ix)+i=0$ or

$(e^{i x} - e^{- i x}) (1 + e^{i x} + e^{- i x}) + i (e^{i x} + e^{- i x} + 1) = 0$ $(e^(ix)-e^(-ix))(1+e^(ix)+e^(-ix))+i(e^(ix)+e^(-ix)+1) = 0$ or

$(e^{i x} + e^{- i x} + 1) (e^{i x} - e^{- i x} + i) = 0$ $(e^(ix)+e^(-ix)+1)(e^(ix)-e^(-ix)+i)=0$ or

$(2 cos x + 1) (2 sin x + 1) = 0$ $(2cosx+1)(2sin x+1)=0$

so the solutions are

$cos x = - \frac{1}{2}$ $cosx = -1/2$ and $sin x = - \frac{1}{2}$ $sin x = -1/2$ or

$(x = \frac{2}{3} π + 2 k π) \cup (x = \frac{7}{6} π + 2 k π)$ $(x = 2/3pi+2kpi) uu (x = 7/6pi+2kpi)$ for $k = 0, \pm 1, \pm 2, \dots$ $k = 0,pm1,pm2,cdots$

Answer 2 · 2016-09-25T15:54:05

$2 sin x + 2 cos x + 2 sin 2 x + 1 = 0$ $2sinx + 2cosx + 2sin2x + 1 = 0$

$2 sin x + 2 cos x + 2 (2 sin x cos x) + 1 = 0$ $2sinx + 2cosx + 2(2sinxcosx) + 1= 0$

$2 sin x + 2 cos x + 4 sin x cos x + 1 = 0$ $2sinx + 2cosx + 4sinxcosx + 1 = 0$

${(2 sin x + 2 cos x)}^{2} = {(- 1 - 4 sin x cos x)}^{2}$ $(2sinx + 2cosx)^2 = (-1 - 4sinxcosx)^2$

$4 {sin}^{2} x + 4 {cos}^{2} x + 8 sin x cos x = 1 + 8 sin x cos x + 16 {sin}^{2} x {cos}^{2} x$ $4sin^2x + 4cos^2x + 8sinxcosx = 1 + 8sinxcosx + 16sin^2xcos^2x$

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

$4 - 1 - 16 {sin}^{2} x {cos}^{2} x = 0$ $4 - 1 - 16sin^2xcos^2x = 0$

$3 = 16 {sin}^{2} x (1 - {sin}^{2} x)$ $3 = 16sin^2x(1 - sin^2x)$

$3 = 16 {sin}^{2} x - 16 {sin}^{4} x$ $3 = 16sin^2x - 16sin^4x$

$0 = - 16 {sin}^{4} x + 16 {sin}^{2} x - 3$ $0 = -16sin^4x + 16sin^2x - 3$

$0 = - 16 {sin}^{4} x + 4 {sin}^{2} x + 12 {sin}^{2} x - 3$ $0 = -16sin^4x + 4sin^2x + 12sin^2x - 3$

$0 = - 4 {sin}^{2} x (4 {sin}^{2} x - 1) + 3 (4 {sin}^{2} x - 1)$ $0 = -4sin^2x(4sin^2x - 1) + 3(4sin^2x - 1)$

$0 = (- 4 {sin}^{2} x + 3) (4 {sin}^{2} x - 1)$ $0= (-4sin^2x + 3)(4sin^2x- 1)$

$sin x = \pm \frac{\sqrt{3}}{2} AND sin x = \pm \frac{1}{2}$ $sinx= +- sqrt(3)/2" AND "sinx = +-1/2$

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

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 · 2016-09-25T17:57:00

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

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

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

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

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

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

Again when

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

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3 Answers

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