Question #7c3e7

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Answer 1 · 2017-02-20T14:30:39

#sin15#

#=sin(60-45)#

#=sin60cos45-cos60sin45#

#=sqrt3/2*1/sqrt2-1/2*1/sqrt2#

#=(sqrt3-1)/(2sqrt2)#

#cos15#

#=cos(60-45)#

#=cos60cos45+sin60sin45#

#=1/2*1/sqrt2+sqrt3/2*1/sqrt2#

#=(sqrt3+1)/(2sqrt2)#

Now given equation

#(sqrt3-1)cosx+(sqrt3+1)sinx=2#

Dividing both sides by #2sqrt2# we get

#(sqrt3-1)/(2sqrt2)cosx+(sqrt3+1)/(2sqrt2)sinx=1/sqrt2#

#=>sin15^@cosx+cos15^@sinx=1/sqrt2#

#=>sin(x+15^@)=sin45^@#

#=>sin(x+pi/12)=sin(pi/4)#

#=>x+pi/12=npi+(-1)^npi/4" where " n in ZZ#

#=>x=npi+(-1)^npi/4-pi/12" where " n in ZZ#