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=pix=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 xx. In this case it is easiest to use cosxcosx.

First, we can use the Pythagorean identity sin^2x+cos^2x=1sin2x+cos2x=1 rearranged as sin^2x=1-cos^2xsin2x=1−cos2x and plug it into the equation.

1-cos^2x+cosx=-11−cos2x+cosx=−1. This is actually a quadratic in cos xcosx!

Let u=cosxu=cosx, our equation is now 1-u^2+u=-11−u2+u=−1.

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

Since u=cosxu=cosx, we now know that cosx=-1cosx=−1 or cosx=2cosx=2. This second case is impossible, since the range of cosxcosx is -1<=y<=1−1≤y≤1, which does not include 22.

cosx=-1cosx=−1 so x=arccos(-1)x=arccos(−1), therefore on the interval 0<=x<2pi0≤x<2π, x=pix=π.

Nghi N
Dec 8, 2017

0; pi/2; (2pi)/30;π2;2π3

Explanation:

sin^2 x + cos x = 1sin2x+cosx=1
Replace sin^2 x by (1 - cos^2 x)(1−cos2x) -->
1 - cos^2 x + cos x = 11−cos2x+cosx=1
- cos^2 x + cos x = 0−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 = pi/2x=π2, and x = (3pi)/2x=3π2
b. - cos x + 1 = 0 --> cos x = 1
x = 0 and x = 2pix=2π.
Answers for interval [0, 2pi)[0,2π):
0; pi/2; (2pi)/3;0;π2;2π3;
Note: 2pi is not included in the solution set --> 0 <= x < 2pi0≤x<2π

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