How do you find all solutions of the equation in the interval [0, 2pi) for this
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How do you find all solutions of the equation in the interval [0, 2pi) for this

[From: ] [author: ] [Date: 12-08-02] [Hit: ]
Let y = sin x, then 2y^2 + 3y + 1 = 0. Its a quadratic equation which you solve with standard methods.-that equation creates a curve on a graph, in other words there are an infinite number of solutions. I would just graph it.......
2sin^2x + 3sinx + 1 = 0

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

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

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

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

sin(x) = -1/2 => x = 7π/6, 11π/6

sin(x) = -1 => x = 3π/2

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Let's say you find the value of sin x. Find one value of x with that sine. Then (pi - x) has the same sine.

Since it's a quadratic, there might be two values of sin x that are solutions. You have to do that for each one.

Now, how do you find the value of sin x? Let y = sin x, then 2y^2 + 3y + 1 = 0. It's a quadratic equation which you solve with standard methods.

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that equation creates a curve on a graph, in other words there are an infinite number of solutions. I would just graph it.
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