Question

Help please!!!!
Find the exact zeroes of the function in the interval [0,2pi):
f(x) = sin x/2 - cos x

Answer 1

use the half angle formula...

Answer 2

\[\sin \frac{x}{2} - cosx = 0\]

Answer 3

\[\sin \frac{x}{2} = \pm \sqrt{\frac{1-cosx}{2}} \]

Answer 4

now your equation can be written as:\[\pm \sqrt{\frac{1-cosx}{2}} = cosx\]

Answer 5

squaring both sides gives:
(1-cosx)/2 = cos^2(x)

2cos^2(x) + cosx -1 = 0
notice this is a trig equation in quadratic form so see if you can factor:
(2cosx - 1)(cosx + 1) = 0

equate each factor to 0 and solve. ...should be easy from here.