Question #ac6f8

2 Answers
Nghi N.
Aug 12, 2015

Solve: #cos^2 (pi/6) - sin^2 (pi/6 - x) = 0#

#x = - pi/6# and #x = pi/2#

Explanation:

#[cos (pi/6) - sin (pi/6 - x)][cos (pi/6) + sin (pi/6 - x)] # = 0

a. #cos (pi/6) - sin (pi/6 - x) = 0#
Replace #sin (pi/6 - x)# by #cos (pi/2 - pi/6 + x)# we get:
#cos (pi/6) = cos (pi/3 + x)#
#pi/6 = +- (pi/3 + x) #
#x = pi/6 - pi/3 = -pi/6#
#x = pi/6 + pi/3 = pi/2#
b. #cos (pi/6) + sin (pi/6 - x) = 0#
#cos (pi/6) = -sin (pi/6 + x) = cos (pi/2 - pi/6 - x) = cos (pi/3 - x)#
#pi/6 = +- (pi/3 - x)#
#pi/6 = pi/3 - x #--> #x = pi/6 - pi/3 = -pi/6#
#pi/6 = -pi/3 + x --> x = pi/6 + pi/3 = 3pi/6 = pi/2#
Finally, within interval #(-pi/2, pi/2)#, two answers:
#-pi/6# and #pi/2#
Check. #x = -pi/6# --> #cos^2 pi/6 = 0.75# --> #sin^2 (pi/6 + pi/6) =, sin^2 pi/3 = 0.75#. Then 0.75 - 0.75 = 0. OK

mason m
Oct 2, 2016

#(2cos^2(x)+sqrt3sin(2x))/4#

Explanation:

We cannot solve this, since it is not an equation, but we can simplify it.

Recall that #sin(A-B)=sin(A)cos(B)-cos(A)sin(B)#.

So, first, we see that

#sin(pi/6-x)=sin(pi/6)cos(x)-cos(pi/6)sin(x)#

#=1/2cos(x)-sqrt3/2sin(x)#

#=(cos(x)-sqrt3sin(x))/2#

Also, note that #cos^2(pi/6)=(sqrt3/2)^2=3/4#.

Thus:

#cos^2(pi/6)-sin^2(pi/6-x)=3/4-((cos(x)-sqrt3sin(x))/2)^2#

#=(3-(cos(x)-sqrt3sin(x))^2)/4#

#=(3-(cos^2(x)-2sqrt3sin(x)cos(x)+3sin^2(x)))/4#

Note that #2sin(x)cos(x)=sin(2x)#:

#=(3-cos^2(x)+sqrt3sin(2x)-3sin^2(x))/4#

We can simplify this many ways, this being just one:

#=(3-3cos^2(x)+2cos^2(x)+sqrt3sin(2x)-3sin^2(x))/4#

#=(3-3(cos^2(x)+sin^2(x))+2cos^2(x)+sqrt3sin(2x))/4#

#=(3-3+2cos^2(x)+sqrt3sin(2x))/4#

#=(2cos^2(x)+sqrt3sin(2x))/4#

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