Question #bb3f7

1 Answer
Jan 17, 2016

Answer:

#x=pi/2,pi#

Explanation:

The first step is to rewrite #cos(2x)# in terms of #sin(x)#.

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

Thus, the function can be rewritten as

#1-2sin^2(x)+2sin(x)-3=-2#

Move all the terms to the same side.

#0=2sin^2(x)-2sin(x)#

Divide both sides by #2#.

#sin^2(x)-sin(x)=0#

Factor a #sin(x)# term.

#sin(x)(sin(x)-1)=0#

Here, we have a product of two terms that equals #0#. This means that either one of the two terms could be equal to #0#:

#sinx=0color(white)(ssssss)"or"color(white)(ssssss)sin(x)-1=0=>sin(x)=1#
#x=picolor(white)(ssssssskl)"or"color(white)(ssssssssssssssssssssssss)x=pi/2#

The times when #sin(x)=0,1# for #0 < x < 2pi# are #x=pi/2,pi#.