How do you use the half-angle identities to find all solutions on the interval [0,2pi) for the equation #sin^2x = cos^2 (x/2)#?

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Answer 1 · 2016-07-12T10:09:12

#pi/3, pi and (5pi)/3. #

Use #cos 2a = 2 cos^2 a -1#.

The given equation is

#sin^2 x=1-cos^2 x=1-(2cos^2 (x/2)-1)^2= cos^2 (x/2)#.

Rearranging and simplifying,

# cos^2(x/2)(4 cos^2(x/2)+3)=0#

The second factor = 0 gives solutions, with #cos (x/2)=+-sqrt 3/2>1#.

So, these give # x/2= pi/6 and x/2=(5pi)/6 in (0, pi), and so,

#x=pi/3 and (5pi)/3 in (0, 2pi). inadmissible.solutions.

Also, the first factor #cos^2 (x/2)=0#, gives #x/2 = pi/2 in (0, pi)#, and so,

#x=pi in (0, 2pi)#