Question #4a4ea

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Answer 1 · 2018-01-20T01:51:23

#x=pi/2#

#x=(3pi)/2#

.

#cos^2x-cos2x=1#

We have an identity that says:

#cos^2x=(1+cos2x)/2#

We can substitute it:

#(1+cos2x)/2-cos2x=1#

#(1+cos2x-2cos2x)/2=1#

#(1-cos2x)/2=1#

We also have another identity that says:

#sin^2x=(1-cos2x)/2#

We can substitute it:

#sin^2x=1#

#sinx=+-1#

#sinx=1# gives us #x=pi/2#

#sinx=-1# gives us #x=(3pi)/2#