How can this be solved?

#sin(x/2) + cosx -1 = 0#

1 Answer
Feb 28, 2018

#x=0,pi/3,(5pi)/3#

Explanation:

Use the half-angle formula for sine:

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

First, use this formula. Then, isolate the radical and square both sides of the equation. Lastly, create a quadratic out of #cos(x)#'s and solve:

#sin(x/2)+cosx-1=0#

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

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

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

#(1-cosx)/2=cosx^2-2cosx+1#

#1-cosx=2cos^2x-4cosx+2#

#0=2cos^2x-3cosx+1#

#0=(2cosx-1)(cosx-1)#

Solve for when each factor equals #0#:

#color(white){color(black)( (2cosx-1=0, qquadcosx-1=0), (2cosx=1, qquadcosx=1), (cosx=1/2,)):}#

Here's a unit circle to remind us of where these certain cosine values are:

https://en.wikipedia.org/wiki/Unit_circle

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

#cosx=1=>x=0#

So:

#x=0, pi/3, (5pi)/3#

Lastly, since we squared both sides of the equation earlier, there is a possibility that new answers that don't actually work were added. That means that we have to test all the answers we got in the original problem:

#sin(x/2)+cosx-1=0#

Testing #0#:

#sin(0/2)+cos0-1stackrel?=0#

#0+1-1stackrel?=0#

#0=0#

The solution #0# works. Testing #pi/3#:

#sin((pi/3)/2)+cos(pi/3)-1stackrel?=0#

#sin(pi/6)+1/2-1stackrel?=0#

#1/2-1/2stackrel?=0#

#0=0#

The solution #pi/3# works. Testing #(5pi)/3#:

#sin(((5pi)/3)/2)+cos((5pi)/3)-1stackrel?=0#

#sin((5pi)/6)+1/2-1stackrel?=0#

#1/2-1/2stackrel?=0#

#0=0#

All the solutions work. The final answers are:

#x=0,pi/3,(5pi)/3#