Question #e650c

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Answer 1 · 2017-01-27T07:54:50

The solutions are #S={pi, pi/3, 5/3pi}#

We need

#cos2x=1-2sin^2x#

#cosx=1-2sin^2(x/2)#

Therefore,

#2sin^2(x/2)=1-cosx#

#sin^2(x/2)=(1-cosx)/2#

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

Our equation is

#sin(x/2)+cosx=0#

Therefore,

#sqrt((1-cosx)/2)+cosx=0#

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

Squaring both sides

#(1-cosx)/2=cos^2x#

#1-cosx=2cos^2x#

The new equation is

#2cos^2x+cosx-1=0#

We solve this like the quadratic equation

#ax^2+bx+c=0#

We start by calculating the discriminant

#Delta=b^2-4ac=1-4*2+(-1)=9#

#Delta>0#, there are 2 real roots

#cosx=(-b+-sqrt(Delta))/(2a)#

#=(-1+-3)/4#

The roots are

#cosx=-1#, #=>#, #x=pi#

#cosx=1/2#, #=>#, #x=pi/3 , 5/3pi#