Solve sin θ/2=-cos θ/2 on the interval [0,2pi)?

Use the sine and cosine half-angle formulas:

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

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

In this case, the plus-or-minus doesn't matter because we're going to be squaring both sides of the equation. No matter which one it is - plus or minus - it will be positive after squaring.

Here's the problem:

#sin(theta/2)=-cos(theta/2)#

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

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

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

#1-cosx=1+cosx#

#-cosx=cosx#

#-2cosx=0#

#cosx=0#

Here's a unit circle to remind us where #cosx=0#:

The only places where #cosx=0# is:

#x=pi/2, (3pi)/2#

These are the solutions for now.

Since we squared both sides of the equation, we're going to have to check our answers since we may have unintentionally added some that don't actually work.

Checking #pi/2#:

#sin((pi/2)/2)stackrel?=-cos((pi/2)/2)#

#sin(pi/4)stackrel?=-cos(pi/4)#

#sqrt2/2!=-sqrt2/2#

The solution #pi/2# actually doesn't work. Now checking #(3pi)/2#:

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

#sin((3pi)/4)stackrel?=-cos((3pi)/4)#

#sqrt2/2stackrel?=-(-sqrt2/2)#

#sqrt2/2=sqrt2/2#

The solution #(3pi)/2# does work, so that's the only answer. Hope this helped!