How do you solve #cos2x = sinx + cosx# on the interval [0,2pi]?

2 Answers
Mar 17, 2016

#x = 0,135,315,90,180#

Explanation:

#cos 2x = sin x + cos x#

#Cos^2 x - sin^2 x = cos x + sin x#

#(cosx + sinx)(cosx - sinx) = cosx + sin x#

#cancel((cosx + sinx))(cosx - sinx) =cancel( (cosx + sin x))#

#cos x - sin x = 1#

Square both sides

#(cos x - sin x)^2 = 1#

#cos ^2 x + sin ^2 x -2sinx cosx = 1#

#cancel1 - 2sinx cos x = cancel1 #

#sin 2x = 0#

#=> 2x = 0=> x = 0#

Another method

#cos 2x - sin x - cos x = 0#

#cos^2x - sin^2x -(sinx + cosx) = 0#

#(Cosx - sinx)(cosx + sinx) -(sinx + cosx) = 0#

#color(blue)((sinx + cosx))color(red)((cosx - sinx -1)) = 0#

#color(blue)("Blue part")#

#color(blue)((sinx + cosx)) = 0#

#color(blue)(sinx = - cos x)#

#tanx = -1#

# x = 135,315#

#color(red)("red part") #

#color(red)((cosx - sinx -1)) = 0#

#color(red)(cos x - sin x = 1#

Square both sides

#(cos x - sin x)^2 = 1#

#cos ^2 x + sin ^2 x -2sinx cosx = 1#

#cancel1 - 2sinx cos x = cancel1 #

#sin 2x = 0#

#=> 2x = 0,180,360=> x = 0,90,180#

#0, 5pi/4,, 3pi/2, 7pi/4 and 2pi#..

Explanation:

cos 2x = #cos^2x-sin^2x# = #(cos x - sin x)(cos x + sin x)#

The given equation is
#(cos x - sin x-1)(cos x + sin x) = 0#

Solve separately #cos x - sin x-1 = 0# and #cos x + sin x = 0#.

The solutions for the first are x = 0, #3pi/2, 2pi#.
For the other, x = #3pi/4, 7pi/4#.