Thread: 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?

review your unit circle ...

at and

You are right to discard Sin x = 2. But there are two solutions for Sin x = -1/2.

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.

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?