Find all solutions of each equation in the interval [0,2pi) (sinx+1)(tanx-1)=0

Question

Find all solutions of each equation in the interval [0,2π)

(sinx+1)(tanx-1)=0

why is 3π/2 not part of the final solution set?

Answer 1

Answer 2

f(x) = (sin(x) + 1)(tan(x) - 1) = 0 ==> (sin(x) + 1)([sin(x)/cos(x)] - 1) = 0

When a product of factors = 0, we must explore the possibility of each factor being 0, and in this case, keep an eye on that cos(x) in the denominator.

a. sin(x) + 1 = 0 ==> sin(x) = -1 ==> x = 270º = 3π/2

# 3π/2 is problematic due to the other factor with tan(x) = sin(x)/cos(x) because cos(3π/2) = 0

b. tan(x) - 1 = 0 ==> tan(x) = sin(x)/cos(x) = 1 = 1 / 1 or -1 / -1 ==> x = 45º = π/4 (Quadrant I) or 225º = (180º + 45º) = π + π/4 = 5π/4

Ans. x = π/4 , 5π/4