How would you go about solving sin(x+pi/3) - cos(x+pi/6) = sqrt(2)/2 for all solutions on the interval [0,2pi)?

1 Answer
AJ
Apr 18, 2018

#x=pi/4#

Explanation:

Given,
#sin (x+pi/3)-cos (x+pi/6)=sqrt (2)/2#

Go on expanding
#sinxcos(pi/3)+sin (pi/3)cosx-[cosxcos (pi/6)-sinxsin (pi/6)=sqrt (2)/2#
#=sinx.(1/2)+cosx. sqrt (3)/2-cosx.sqrt (3)/2+six. (1/2)=sqrt (2)/2#
#1/2sinx+cancel(cosxsqrt (3)/2)- cancel(cosxsqrt (3)/2)+1/2sinx=sqrt (2)/2#
#1/2sinx+1/2sinx=sqrt (2)/2#
#sinx=sqrt (2)/2#
#x=arcsin [sqrt (2)/2]#
#:.x=pi/4+2kpi#
For interval #[0,2pi] only one solution exists. Thank you..

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