# Can someone explain this pre-calculus/trigonometry math?

Solve for x, 0<x<2pi

tan x cos x - tan x = 0

i need help asap and please explain how you canceled stuff out and got numbers

Tan(x)[cos(x)] - tan(x) = 0

Take out common term tan(x).

tan(x)[cos(x) - 1] = 0

tan(x) = 0 or cos(x) - 1 = 0

Take tan(x) = 0

The trigonometric table in tan(0) = 0

tan(x) = tan(0)

Apply arc(tan) each side

arc(tan)[tan(x)] = arc(tan)[tan(0)]

x = 0

Take cos(x) - 1 = 0

cos(x) = 1

The trigonometric table in cos(0) = 1

cos(x) = cos(0)

Apply arc(cos) each side

arc(cos)[cos(x)] = arc(cos)[cos(0)]

x = 0

Therefore x = 0.

0 < x < 2π

tanxcosx - tanx = 0

Recall : Trigonometry quotient identities tanx = sinx / cosx

Substitute tanx = sinx / cosx

(sinx / cosx) ×cosx - sinx / cosx = 0

sinx - sinx / cosx = 0

sinx - sinx×secx = 0

sinx(1 - secx) = 0

sinx = 0 or 1 - secx =0

sinx = 0

sinx = sin0 = sinπ (0< x < 2π)

Therefore x = 0 , π

1 - secx = 0

1 = secx

1 = 1 / cosx

cosx = 1 = cos0 (0 < x < 2π)

x = 0

Therefore x = 0, π (0 < x < 2π).