How do you use the inverse functions where needed to find all solutions of the equation $2sin^2x-7sinx+3=0$ in the interval $[0,2pi)$ ?

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How do you use the inverse functions where needed to find all solutions of the equation $2sin^2x-7sinx+3=0$ in the interval $[0,2pi)$ ?

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Answer 1 · 2017-03-11T06:15:46

$pi/6; (5pi)/6$

Explanation:

Solve this quadratic equation for sin x by using the improved quadratic formula (Socratic Search):
$2sin^2 x - 7sin x + 3 = 0$
$D = d^2 = b^2 - 4ac = 49 - 24 = 25$ --> $d = +- 5$
There are 2 real roots:
$sin x = -b/(2a) +- d/(2a) = 7/4 +- 5/4$
a. sin x = 12/4 = 3 (rejected because > 1)
b. $sin x = (7 - 5)/4 = 2/4 = 1/2$
Trig table and unit circle give 2 solutions for $(0, 2pi)$
$x = pi/6$ , and $x = pi - pi/6 = (5pi)/6$

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How do you use the inverse functions where needed to find all solutions of the equation $2sin^2x-7sinx+3=0$ in the interval $[0,2pi)$ ?

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