# math

Find all solutions in the interval [0,2π) that satisfy tan(x)sin²(x) = tan(x) .

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1. tan(x)sin²(x) = tan(x)
tan(x)sin²(x) - tan(x) = 0
tanx(sin² x - 1) = 0
tanx = 0 or sin² x = 1

if tanx = 0, x = 0, 180°, or 360° ---> x = 0,π,2π
if sin² x = 1
sinx = ± 1
x = 45°, 135°, 225° or 315° ----> x = π/4, 3π/4, 5π/4, 7π/4

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2. tan ( x ) ∙ sin²( x ) = tan( x )

Subtract tan ( x ) to both sides

tan ( x ) ∙ sin²( x ) - tan( x ) = 0

tan ( x ) ∙ [ sin²( x ) - 1 ] = 0

Multiply both sides by - 1

tan ( x ) ∙ [ 1 - sin²( x ) ] = 0

tan ( x ) ∙ cos²( x ) = 0

sin ( x ) / cos ∙ cos²( x ) = 0

sin ( x ) ∙ cos ( x ) = 0

( 1 / 2 ) sin ( 2 x ) = 0

Multiply both sides by 2

sin ( 2 x ) = 0

2 x = n ∙ π

x = n ∙ π / 2

n = 0 , ± 1 , ± 2 , ± 3 , ± 4...

For n = 0

x = 0 ∙ π / 2 = 0

For n = 1

x = 1 ∙ π / 2 = π / 2

For n = 2

x = 2 ∙ π / 2 = π

For n = 3

x = 3 ∙ π / 2 = 3 π / 2

For n = 4

x = 4 ∙ π / 2 = 2 π

Interval [ 0 , 2 π ) is a half-open interval.

[ 0 , 2 π ) means greater or equal than 0 and less than 2 π.

So x = 2 π isn't solution.

The solutions are:

x = 0

x = π / 2

x = π

x = 3 π / 2

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posted by Bosnian

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