Find all solutions in the interval [0,2π) that satisfy tan(x)sin²(x) = tan(x) .
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
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
To find all the solutions that satisfy the given equation, tan(x)sin²(x) = tan(x), in the interval [0,2π), you can follow the steps below:
Step 1: Identify any restrictions.
Since the equation involves the tangent function, we need to account for the restrictions on the tangent function. The tangent function is undefined when the angle is π/2 + nπ, where n is an integer. So, we need to exclude these values from our interval.
Step 2: Rewrite the equation.
tan(x)sin²(x) = tan(x)
Divide both sides by tan(x):
sin²(x) = 1
Step 3: Solve the rewritten equation.
Taking the square root of both sides, we have:
sin(x) = ±1
Step 4: Find the values of x that satisfy sin(x) = 1 or sin(x) = -1.
Considering the interval [0,2π):
For sin(x) = 1, the solutions are x = π/2 and x = 3π/2.
For sin(x) = -1, the solutions are x = π/2 + π and x = 3π/2 + π, which simplify to x = 3π/2 and x = 5π/2.
Step 5: Exclude the restricted values.
We need to exclude the values that make the tangent function undefined from our solutions. Thus, we exclude x = π/2, x = 3π/2, x = 3π/2 + π, and x = 5π/2.
Therefore, the solutions that satisfy the given equation in the interval [0,2π) are x = π/2 and x = 5π/2.