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Answer 1 · 2017-02-02T08:22:32

The solutions are #S={-pi/6 , -5/6pi}#

We solve this equation like a quadratic equation

#ax^2+bx+c=0#

#4sin^2x+4sinx+1=0#

Let's calculate the discriminant

#Delta=b^2-4ac=4^4-4*4*1=16-16=0#

As #Delta =0#, there is one double real root

#x=-b/2a#

#sinx=-4/(2*4)=-1/2#

Therefore,

#x=-pi/6 ; -5pi/6#

Answer 2 · 2017-02-02T15:47:13

#4sin^2x+4sinx+1=0#

#=>(2sinx)^2+2*2sinx*1+1^2=0#

#=>(2sinx+1)^2=0#

#=>sinx=-1/2=sin(-pi/6)#

General solution

#x=npi-(-1)^npi/6" where " n in ZZ#

We are to find out x for x# in [-pi;4pi]#

Putting #n=-1#

#x=-pi+pi/6=-(5pi)/6#

Putting #n=0#

#x=-pi/6 in [-pi;4pi]#

Putting #n=1#

#x=pi-(-pi/6)=(7pi)/6 in [-pi;4pi]#

Putting #n=2#

#x=2pi-pi/6=(11pi)/6 in[-pi;4pi]#

Putting#n=3#

#x=3pi-(-pi/6)=(19pi)/6 in [-pi;4pi]#

Putting #n=4#

#x=4pi-(+pi/6)=(23pi)/6 in [-pi;4pi]#