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
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.
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.