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 $[0, 2 pi]$ .

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