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

2 Answers
Dec 7, 2017

#x=pi#

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 #sin^2x+cos^2x=1# rearranged as #sin^2x=1-cos^2x# and plug it into the equation.

#1-cos^2x+cosx=-1#. This is actually a quadratic in #cos x#!

Let #u=cosx#, our equation is now #1-u^2+u=-1#.

Rearranging this into quadratic form we get #u^2-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<2pi#, #x=pi#.

Dec 8, 2017

#0; pi/2; (2pi)/3#

Explanation:

#sin^2 x + cos x = 1#
Replace sin^2 x by #(1 - cos^2 x)# -->
#1 - cos^2 x + cos x = 1#
#- cos^2 x + cos x = 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/2#, and #x = (3pi)/2#
b. - cos x + 1 = 0 --> cos x = 1
x = 0 and #x = 2pi#.
Answers for interval #[0, 2pi)#:
#0; pi/2; (2pi)/3;#
Note: 2pi is not included in the solution set --> # 0 <= x < 2pi#