Solve, given that (0^{o} < x < 360^{o}): 2 sin^{2} x - 3 sin x + 1 = 0

I have to solve for x, and x has to be an angle.

Shades
Jan 6, 2016

#1**+5 **

Solve for x:

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

The left hand side factors into a product with two terms:

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

Split into two equations:

sin(x)-1 = 0 or 2 sin(x)-1 = 0

Add 1 to both sides:

sin(x) = 1 or 2 sin(x)-1 = 0

Take the inverse sine of both sides:

x = pi/2+2 pi n_1 for n_1 element Z

or 2 sin(x)-1 = 0

Add 1 to both sides:

x = pi/2+2 pi n_1 for n_1 element Z

or 2 sin(x) = 1

Divide both sides by 2:

x = pi/2+2 pi n_1 for n_1 element Z

or sin(x) = 1/2

Take the inverse sine of both sides:

**Answer: | | x = pi/2+2 pi n_1 for n_1 element Z or x = (5 pi)/6+2 pi n_2 for n_2 element Z or x = pi/6+2 pi n_3 for n_3 element Z**

Guest Jan 6, 2016

#3**0 **

That is because you don't understand "complex solutions"!!. If you want to go to college and study advanced Math, your better start honing up on it, because it is the only game in town at that level.

Guest Jan 6, 2016

#4**0 **

Okay, someone please get an admin in here... Guest, I know how to do math, this problem is one that I do all of the time, I just can't figure this one out.

Shades
Jan 6, 2016

#5**+5 **

Hi Shades :)

2 sin2 x - 3 sin x + 1 = 0

\(2 sin^2 x - 3 sin x + 1 = 0\\ let\;\; y=sin^2x\\ 2y^2-3y+1=0\\ 2y^2-2y-y+1=0\\ 2y(y-1)-1(y-1)=0\\ (2y-1)(y-1)=0\\ 2y-1=0 \qquad or \qquad y-1=0\\ 2y=1 \qquad or \qquad y=1\\ y=\frac{1}{2} \qquad or \qquad y=1\\ so\\ sin^2x=\frac{1}{2} \qquad or \qquad sin^2x=1\\ sinx=\pm \frac{1}{\sqrt{2}} \qquad or \qquad sinx=\pm1\\\)

\(x=45^0,135^0,225^0,315^0, 90^0,\;\;or \;\;270^0\)

Melody
Jan 6, 2016