How do you solve $cos 2 x + 3 sin x - 2 = 0$ $cos 2x + 3 sinx - 2= 0$ ?

Question

23176 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-18T16:09:35

$S = {\frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, x = \frac{π}{2} + 2 π n}$ $S={pi/6 +2pin, (5pi)/6 +2pin, x=pi/2+2pin}$

Use Double Argument Property:
$cos 2 A = 1 - 2 {sin}^{2} A$ $cos2A=1-2sin^2A$

$1 - 2 {sin}^{2} x + 3 sin x - 2 = 0$ $1-2sin^2x+3sinx-2=0$

$2 {sin}^{2} x - 3 sin x + 1 = 0$ $2sin^2x-3sinx+1=0$

$(2 sin x - 1) (sin x - 1) = 0$ $(2sinx-1)(sinx-1)=0$

$2 sin x - 1 = 0 or sin x - 1 = 0$ $2sinx-1=0 or sinx-1=0$

$sin x = \frac{1}{2} or sin x = 1$ $sinx=1/2 or sinx =1$

$x = {sin}^{- 1} (\frac{1}{2}) or x = {sin}^{- 1} 1$ $x=sin^-1(1/2) or x=sin^-1 1$

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n or x = \frac{π}{2} + 2 π n$ $x=pi/6 +2pin, (5pi)/6 +2pin or x=pi/2+2pin$

$S = {\frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, x = \frac{π}{2} + 2 π n}$ $S={pi/6 +2pin, (5pi)/6 +2pin, x=pi/2+2pin}$

How do you solve cos 2x + 3 sinx - 2= 0cos2x+3sinx−2=0cos 2x + 3 sinx - 2= 0?