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

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

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