Solve 2 sin(3\theta ) - 1 = 0

Evaluate

2 sin (3 θ) - 1 = 0

Move the constant to the right-hand side and change its sign

2 sin (3 θ) = 0 + 1

Removing 0 doesn't change the value,so remove it from the expression

2 sin (3 θ) = 1

Divide both sides

\frac{2 sin ( 3 θ )}{2} = \frac{1}{2}

Divide the numbers

sin (3 θ) = \frac{1}{2}

Use the inverse trigonometric function

3 θ = arcsin (\frac{1}{2})

Calculate

3 θ = \frac{π}{6} 3 θ = \frac{5 π}{6}

Add the period of 2 kπ, k \in Z to find all solutions

3 θ = \frac{π}{6} + 2 kπ, k \in Z 3 θ = \frac{5 π}{6} + 2 kπ, k \in Z

Calculate

More Steps

θ = \frac{π}{18} + \frac{2 kπ}{3}, k \in Z 3 θ = \frac{5 π}{6} + 2 kπ, k \in Z

Calculate

More Steps

θ = \frac{π}{18} + \frac{2 kπ}{3}, k \in Z θ = \frac{5 π}{18} + \frac{2 kπ}{3}, k \in Z

Solution

θ = {\frac{π}{18} + \frac{2 kπ}{3} \frac{5 π}{18} + \frac{2 kπ}{3}, k \in Z

Alternative Form

θ = {1 0^{\circ} + 12 0^{\circ} k 5 0^{\circ} + 12 0^{\circ} k, k \in Z

Alternative Form

θ \approx {0.174533 + \frac{2 kπ}{3} 0.872665 + \frac{2 kπ}{3}, k \in Z

Solve the equation