find the solutons of sinx^2+2sinx-2=0

Question

I need a step by step solution

Answer 1

 

sin²x+2sinx-2=0

let n=sinx

n²+2n-2=0

use the quadratic equation

a=1, b=2, c=-2

[-2±√(4+8)]/2

[-2±√12]/2

[-2±2√3]/2

-1±√3

n=-1+√3=-1+1.73205=0.73205

n=-1-√3=-1-1.73205=-2.73205

n=sinx

sinx=0.73205

using arcsin calculator or a table or...

sin(47.05852922º)

x=47.05852922º or 47º 3' 30.705"

sin is not defined for -2.73205

Forgot about second quadrant. Look at Mark M.'s answer in radians. I'll add my correction in degrees.

180º-47.05852922º=132.9414708º

Answer 2

sin²x + 2sinx - 2 = 0

 

Let u = sinx.  Then the equation can be written as u²+2u-2=0.

 

By the quadratic formula, u = (-2±√12)/2 = 0.732 or -2.732

 

So, sinx = 0.732 or -2.732

 

Since sinx is always in the interval [-1,1], -2.732 can be eliminated.  

 

So, sinx = 0.732

 

sinx is positive in quadrants 1 and 2

 

x = sin^-1(0.732) or π-sin^-1(0.732)

 

   ≈ 0.82 or 2.32 (answers in radians)