precalc math problem 12

Question

Solve the equation on the interval [0,2π).

sin2x=− square root of 3 sinx

Only the 3 is in a square root

Answer 1

sin(2x) = -√3(sinx)

2sinxcosx + (√3)sinx = 0

sinx(2cosx + √3) = 0

sinx = 0 or cosx = -√3/2

sinx = 0 when x = 0 or π

cosx = -√3/2 when x = 5π/6 or 7π/6

Answer 2

sin2x = -(sqr3)sinx

2sinxcosx = -(sqr3)sinx

2sinxcosx + (sqr3)sinx = 0

sinx(2cosx +sqr3) = 0

set each factor = 0

sinx = 0, x =0 or pi

2cosx =-sqr3

cosx = -sqr3)/2

x = 7pi/6 or 5pi/6

four solutions:

x = 0, 5pi/6, pi, and 7pi/6