How do you find all real numbers on the interval [0,2pi) that satisfy the equation and use radian measure $2sin^2(x)-3sin(x)=-1$ ?

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How do you find all real numbers on the interval [0,2pi) that satisfy the equation and use radian measure $2sin^2(x)-3sin(x)=-1$ ?

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Answer 1 · 2016-03-24T13:49:26

$pi/6, pi/2, (5pi)/6$

Explanation:

$f(x) = 2sin^2 x - 3sin x + 1 = 0.$
Solve this quadratic equation for sin x.
Since a + b + c = 0, use shortcut. The 2 real roots are:
sin x = 1 and sin x = c/a = 1/2.
a. sin x = 1 --> $x = pi/2$
b. $sin x = 1/2.$ The trig unit circle gives 2 arcs that have the same sin value (1/2)
$x = pi/6 and x = pi - pi/6 = (5pi)/6$

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How do you find all real numbers on the interval [0,2pi) that satisfy the equation and use radian measure $2sin^2(x)-3sin(x)=-1$ ?

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Explanation:

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How do you find all real numbers on the interval [0,2pi) that satisfy the equation and use radian measure $2sin^2(x)-3sin(x)=-1$ ?