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

##### 1 Answer

Feb 9, 2016

#### Explanation:

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^2 - u - 2 = 0#

This quadratic equation has two solutions:

Resubstitute:

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

Graph of

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