How do you solve the following equation #-2sin^2x=sin x-1# in the interval [0, 2pi]?

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How do you solve the following equation #-2sin^2x=sin x-1# in the interval [0, 2pi]?

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Answer 1 · 2016-10-11T03:56:49

#pi/6; (5pi)/6; (3pi)/2#

Explanation:

Bring the quadratic equation to standard form, then solve it for sin x:
#2sin^2 x + sin x - 1 = 0#
Since a - b + c = 0, use shortcut. One real root is sin x = -1 and the other is #sin x = -c/a = 1/2#
Use trig table and unit circle:
a. sin x = -1 --> arc #x = (3pi)/2#
b. #sin x = 1/2# -->
arc #x = pi/6# and arc #x = (5pi)/6#

Answers for #(0, 2pi)#
#pi/6; (5pi)/6, (3pi)/2#

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How do you solve the following equation #-2sin^2x=sin x-1# in the interval [0, 2pi]?

1 Answer

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