Find all solutions in the interval [0,2π)

Calculus
Tutor: None Selected Time limit: 1 Day

√2 sin(a/2) + 1 = 0

Mar 17th, 2015

sin(a/2) = +/-sqrt[(1- cosa)/2]

Therefore,

√2 sin(a/2) + 1 = 0

= sqrt(2)(+/-sqrt[(1- cosa)/2]) = - 1

Square both sides to get

2(1- cosa)/2) = 1

= 1 - cosa = 1

- cosa = 0

cosa = 0

a = pi/2, 3pi/2

Mar 17th, 2015

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