Answer:
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
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