Solve for theter in the equation; sin(3theter+120)^o=squareroot(3)/2 for 0^o<theter<180^o

The given equation is sin(3θ + 120°) = √3/2.

Step 1: We know that sin(60°) = √3/2. So, we can set 3θ + 120° = 60°.

Step 2: Solve for θ. Subtract 120° from both sides to get 3θ = -60°.

Step 3: Divide both sides by 3 to solve for θ. θ = -60°/3 = -20°.

However, the given range for θ is 0° < θ < 180°. The solution -20° is not within this range.

Step 4: We also know that sin(180° - α) = sin(α). So, we can also set 3θ + 120° = 180° - 60° = 120°.

Step 5: Solve for θ. Subtract 120° from both sides to get 3θ = 0°.

Step 6: Divide both sides by 3 to solve for θ. θ = 0°/3 = 0°.

Again, this solution is not within the given range for θ.

Step 7: We also know that sin(α + n360°) = sin(α), where n is an integer. So, we can also set 3θ + 120° = 60° + n360°.

Step 8: Solve for θ. Subtract 120° from both sides to get 3θ = -60° + n*360°.

Step 9: Divide both sides by 3 to solve for θ. θ = (-60° + n*360°)/3.

Step 10: Find the values of n