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Question

#sinxcosx + 1/2 =cos^2(x/2 -pi/6)#

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Answer 1 · 2018-03-16T17:42:03

#sinxcosx + 1/2 =cos^2(x/2 -pi/6)#

#=>2sinxcosx + 1 =2cos^2(x/2 -pi/6)#

#=>2sinxcosx + 1 =1+cos(2(x/2 -pi/6))#

#=>sin2x =cos(x -pi/3)#

#=>cos(x -pi/3)=cos(pi/2-2x)#

#=>x-pi/3=2npipm(pi/2-2x)"where "ninZZ#

So

#x-pi/3=2npi+(pi/2-2x)"where "ninZZ#

#=>3x=2npi+pi/2+pi/3"where "ninZZ#

#=>x=2/3npi+(5pi)/18"where "ninZZ#

Again

#x-pi/3=2npi-(pi/2-2x)"where "ninZZ#

#x-pi/3=2npi-pi/2+2x)"where "ninZZ#

#-x=2npi-pi/2+pi/3"where "ninZZ#

#x=-2npi+pi/6"where "ninZZ#

Answer 2 · 2018-03-16T18:01:42

# x in {5pi/18-2pi/3}uu{pi/6-2kpi}, k in ZZ#.

#sinxcosx+1/2=cos^2(x/2-pi/6)#.

#:. 2sinxcosx+1=2cos^2(x/2-pi/6)#.

As, #2sinthetacostheta=sin2theta, &, 1+cos2theta=2cos^2theta,#

#:. sin2x+1=1+cos{2(x/2-pi/6)}, i.e., #

# sin2x=cos(x-pi/3)#.

#:. cos(pi/2-2x)=cos(x-pi/3)...[because, sintheta=cos(pi/2-theta)]#.

Now, #costheta=cosalpha rArr theta=2kpi+-alpha, k in ZZ#.

#:. cos(pi/2-2x)=cos(x-pi/3)#,

#rArr pi/2-2x=2kpi+-(x-pi/3), k in ZZ#.

If, #pi/2-2x=2kpi+(x-pi/3)#, then, #pi/2+pi/3-2kpi=3x, or, #

# x=1/3(5/6pi-2kpi)=5pi/18-2pi/3, k in ZZ#.

If, #pi/2-2x=2kpi-(x-pi/3)#, then, #pi/2-pi/3-2kpi=x, i.e., #

# x=pi/6-2kpi, k in ZZ#.

Altogether, # x in {5pi/18-2pi/3}uu{pi/6-2kpi}, k in ZZ#.