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

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

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Answer 1 · 2017-03-05T01:58:28

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

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