Posted by **Isaac** on Monday, December 5, 2011 at 12:40pm.

Find the exact solutions of the equation in the interval [0,2pi).

sin(x/2)+cos(x)=0

- trig -
**Reiny**, Monday, December 5, 2011 at 1:17pm
recall the half-angel conversions

cos 2A = 1 - 2sin^2 A

so

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

sin(x/2) + 1 - 2sin^2 (x/2) = 0

2sin^2(x/2) - sin(x/2) - 1 = 0

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

sin(x/2) = -1/2 or sin(x/2) = 1

x/2 = 7π/6 or x/2 = 11π/6 or x/2 = π/2

x = 7π/3 or x = 11π/3 or x = π (420° , 660· , 180° )

so for the given domain

x = π

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