Solve the equation in x for exact solutions over the interval [0, 2pi) for sin . x/2=sqrt2-sinx/2 ?

sin#x/2=sqrt(2)-#sin#x/2#

sin#x/2=sqrt(2)-#sin#x/2#

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Feb 9, 2018

Answer:

#color(blue)(90^@, 270^@)#

Explanation:

Using identity:

#color(red)bb(sin^2(theta/2)=1/2(1-costheta)#

#color(red)bb(=> sin(theta/2)=sqrt(1/2(1-costheta))#

#sin(x/2)=sqrt(2)-sin(x/2)#

#sin(x/2)=(sqrt(2))/2#

Substituting:

#sqrt(1/2(1-cosx))=sqrt(2)/2#

Squaring:

#1/2(1-cosx)=2/4#

#1/2-1/2cosx=1/2#

#cosx=0#

#x=arccos(cosx)=arccos(0)=color(blue)(90^@, 270^@)#

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Nghi N Share
Feb 10, 2018

Answer:

#pi/2; (3pi)/2#

Explanation:

#2sin (x/2) = sqrt2#
#sin (x/2) = sqrt2/2#
Trig table and unit circle give 2 solutions -->

a. #x/2 = pi/4#, --> #x = pi/2#.

b. #x/2 = (3pi)/4# --> #x = (3pi)/2#

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