What are the solutions to the equation sin^2x+cosx=-1 on the interval 0 ≤ x < 2pi?

2 Answers
Scott F.
Dec 7, 2017

x=π

Explanation:

To solve an equation like this, it is helpful to have only one trig function so we can solve for it and then solve for x. In this case it is easiest to use cosx.

First, we can use the Pythagorean identity sin2x+cos2x=1 rearranged as sin2x=1−cos2x and plug it into the equation.

1−cos2x+cosx=−1. This is actually a quadratic in cosx!

Let u=cosx, our equation is now 1−u2+u=−1.

Rearranging this into quadratic form we get u2−u−2=0. This can be solved by factoring: (u+1)(u−2)=0 so u=−1 or u=2.

Since u=cosx, we now know that cosx=−1 or cosx=2. This second case is impossible, since the range of cosx is −1≤y≤1, which does not include 2.

cosx=−1 so x=arccos(−1), therefore on the interval 0≤x<2π, x=π.

Nghi N
Dec 8, 2017

0;π2;2π3

Explanation:

sin2x+cosx=1
Replace sin^2 x by (1−cos2x) -->
1−cos2x+cosx=1
−cos2x+cosx=0
(cos x)(- cos x +1) = 0
Either one of the factors should be zero.
a. cos x = 0 --> Unit circle gives 2 solutions:
x=π2, and x=3π2
b. - cos x + 1 = 0 --> cos x = 1
x = 0 and x=2π.
Answers for interval [0,2π):
0;π2;2π3;
Note: 2pi is not included in the solution set --> 0≤x<2π

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