Asked by denise

Find all solutions in the unterval [0,2pi), cos x/2-sinx=0
Answered by Reiny
from sin2A = 2sinAcosA we can say that
sinx = 2sin(x/2)cos(x/2)

cos(x/2) = 2sin(x/2)cos(x/2)
1= 2sin(x/2)
sin(x/2) = 1/2
x/2 = 30°, 150° or π/6, 5π/6 radians
x = 60°, 300° or π/3, 5π/3 radians
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