Solve the equation on the interval 0<x<2pi Cos^2x-sin^2x=1-sinx What is the whole solutions set?

Using double angles I got Cos2x=1-sinx
but don't know what to do next.

1 Answer
Nghi N.
Feb 15, 2016

#0, pi, 2pi, pi/6 and (5pi)/6#

Explanation:

f(x) = cos^2 x - sin^2 x - 1 + sin x = 0
Replace #cos^2 x by (1 - sin^2 x)# -->
#f(x) = 1 - sin^2 x - sin^2 x - 1 + sin x = 0#
#f(x) = -2sin^2 x + sin x = 0#
#f(x) = sin x(-2sin x + 1) = 0#
Two solutions:
#sin x = 0# --> x = 0 and #x = pi# and #x = 2pi#
#sin x = 1/2# --> #x = pi/6# and #x = (5pi)/6#
#x = 0, pi, 2pi, pi/6 and (5pi)/6#

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