Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)
3 tan(θ) sin(θ) − 2 tan(θ) = 0

Note from question, Let K be any integer.  Integer = 1

θ = πk

  θ = 3.142 * 1

  θ = 3.142 in three decimal places

θ = sin⁻¹ (2/3) + 2kπ

   θ = sin⁻¹0.667 + 2*1*3.142

   θ = 0.718 + 6.284

   θ = 7.002 in three decimal places

∴ 7.002 , 3.142

Step-by-step explanation:

Considering the equation

3 tan(θ) sin(θ) − 2 tan(θ) = 0

The objective is to solve the equation.

First solve the equation in one period.

3 tan(θ) sin(θ) − 2 tan(θ) = 0

( 3sinθ − 2 ) tanθ = 0

Therefore, 3sinθ − 2 = 0    also   tanθ = 0

            =>  sinθ = 2/3        ,        tanθ = 0

Pick the right equation.

tanθ = 0

    θ = tan⁻¹ 0

    θ = 0

Using the unit circle, the period of tangent functions is π

Then the general solution of the equation is θ = 0 + πk ==> θ = πk

Pick the left equation.

3sinθ − 2 = 0

3sinθ = 2

sinθ = 2/3

    θ = sin⁻¹ (2/3)

As the sine function has period 2π

Then the general solution is θ = sin⁻¹ (2/3) + 2kπ

answer: i'm not sure

step-by-step explanation:

what is the answer to this