Question
Find the exact solutions of the equation in the interval [0,2pi).
sin(x/2)+cos(x)=0
sin(x/2)+cos(x)=0
Answers
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
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