#38 5.3.2 Precalculus questions

Mathematics
Tutor: None Selected Time limit: 1 Day

May 15th, 2015

2 sin^2 (x) + 3 sin (x) + 1 = 0

sin (x) = t

2t^2+3t+1=0

D=9-2*4*1=1

t1=(-3+1)/4=-1/2   t2=(-3-1)/4=-1

so sin (x) = -1/2 and sin (x) = -1

on the interval [0, 2pi) sin (x) = -1/2 when x = 7pi/6 and x = 11pi/6

on the interval [0, 2pi) sin (x) = -1 when x = 3pi/2

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May 15th, 2015

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