How do you find all solutions of the equation in the interval [0,2pi) given 2sin^2x-3sinx=-1?

1 Answer
Nghi N.
Mar 5, 2017

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

Explanation:

Solve this quadratic equation for sin x:
2sin^2 x - 3sin x + 1 = 0
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 , unit circle gives --> x = pi/2
b. sin x = 1/2 , trig table and unit circle give -->
x = pi/6 and x = pi - pi/6 = (5pi)/6
Answers for (0, 2pi):
pi/6; pi/2, (5pi)/6

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