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

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Answer 1 · 2016-02-09T13:34:34

#x = 3/2 pi#

You can use the law

#sin^2 x + cos^2 x = 1 " "<=> " "cos^2 x = 1 - sin^2 x #

If you plug this into your equation, you have

#sin x = - (1 - sin^2 x) - 1#

#sin x = -2 + sin^2 x#

#sin^2 x - sin x - 2 = 0#

Substitute #u = sin x#:

#u^2 - u - 2 = 0#

This quadratic equation has two solutions: #u = 2# or #u = -1#.

Resubstitute:

#sin x = 2 " "# or #" " sin x = -1#

Graph of #sin x#:

graph{sin(x) [-1.18, 7.588, -2.09, 2.294]}

#sin x = 2# is never true since the range of #sin x# is #[-1, 1]#.

#sin x = -1# is true for #x = 3/2 pi# in the interval #[0, 2 pi]#.