# Solving trignometric equation

• Jul 17th 2012, 01:14 PM
Frederick
Solving trignometric equation
Hi, the problem is:

solve on the interval 0 ≤ x ≤ 2pi

2 sin^2 x - 3 sin x -2 = 0

I factored the equation to (2x + 1)(x - 2)

then separated to 2x + 1 = 0 and x - 2 = 0

finally sin x = -1/2 and sin x = 2

I figured out one of the x's equals 210 degrees or 7pi/6. Sin x = 2 does not seem like a solution. The given choices each have more than one possible solution, and I have only found one. How do I find the other ones?
• Jul 17th 2012, 01:20 PM
skeeter
Re: Solving trignometric equation
review your unit circle ...

$\sin{x} = -\frac{1}{2}$ at $x=\frac{7\pi}{6}$ and $x=\frac{11\pi}{6}$
• Jul 17th 2012, 01:27 PM
Kiwi_Dave
Re: Solving trignometric equation
You are right to discard Sin x = 2. But there are two solutions for Sin x = -1/2.
• Jul 17th 2012, 03:22 PM
HallsofIvy
Re: Solving trignometric equation
Quote:

Originally Posted by Frederick
Hi, the problem is:

solve on the interval 0 ≤ x ≤ 2pi

2 sin^2 x - 3 sin x -2 = 0

I factored the equation to (2x + 1)(x - 2)

then separated to 2x + 1 = 0 and x - 2 = 0

A technical point- since x appears in the orginal equation, it is not a good idea to use "x" to represent "sin(x)". Use some other letter.

Quote:

finally sin x = -1/2 and sin x = 2

I figured out one of the x's equals 210 degrees or 7pi/6. Sin x = 2 does not seem like a solution. The given choices each have more than one possible solution, and I have only found one. How do I find the other ones?