Understanding the Solutions to sin(2x) = -√3/2 on the Interval [0, 2π)
When solving trigonometric equations, the goal is to find all angle measures—usually represented by the variable x—that satisfy the equation within a specific interval. A classic example that yields the solution set x = 2π/3, 5π/6, 5π/3, 11π/6 is the equation sin(2x) = -√3/2 on the interval [0, 2π). This problem serves as an excellent case study for mastering double-angle arguments, reference angles, and the periodic nature of the sine function Most people skip this — try not to..
Deconstructing the Equation: The Double Angle
The presence of 2x instead of x is the defining feature of this problem. It indicates that the argument of the sine function is changing twice as fast as the standard angle x. As a result, the period of sin(2x) is π (since Period = 2π / B, where B=2), rather than the standard 2π.
Because the interval for x is [0, 2π), the interval for the argument 2x expands to [0, 4π). This means we must find all angles θ (where θ = 2x) between 0 and 4π that have a sine value of -√3/2. Once we find those θ values, we simply divide by 2 to retrieve our final solutions for x.
Step 1: Identify the Reference Angle
The absolute value of the ratio is √3/2. From the special right triangles (specifically the 30-60-90 triangle), we know that:
sin(π/3) = √3/2
That's why, the reference angle is π/3 (or 60°). This is the acute angle formed between the terminal side of the angle and the x-axis Not complicated — just consistent. Nothing fancy..
Step 2: Determine the Quadrants
Since the sine value is negative (-√3/2), we look for quadrants where the y-coordinate (sine) is negative. On the unit circle, this occurs in Quadrant III and Quadrant IV.
- Quadrant III Angle: π + Reference Angle = π + π/3 = 4π/3
- Quadrant IV Angle: 2π - Reference Angle = 2π - π/3 = 5π/3
These are the two principal solutions for 2x within the first revolution (0 to 2π).
Step 3: Account for the Extended Interval [0, 4π)
Because our argument is 2x and x goes up to 2π, 2x goes up to 4π. We must find coterminal angles for our principal solutions by adding the period of the sine function (2π) to each.
First Revolution (0 to 2π):
- 2x = 4π/3
- 2x = 5π/3
Second Revolution (2π to 4π): 3. 2x = 4π/3 + 2π = 4π
- 2π = 10π/3
- 2x = 5π/3 + 2π = 11π/3
Now we have four solutions for 2x within the interval [0, 4π): 4π/3, 5π/3, 10π/3, and 11π/3.
Step 4: Solve for x
To find the final solutions for x, we divide each angle by 2:
- x = (4π/3) ÷ 2 = 2π/3
- x = (5π/3) ÷ 2 = 5π/6
- x = (10π/3) ÷ 2 = 5π/3
- x = (11π/3) ÷ 2 = 11π/6
Verification
Let's verify one solution to ensure our method is correct. For x = 5π/6:
sin(2 × 5π/6) = sin(10π/6) = sin(5π/3)
Since 5π/3 is in Quadrant IV where sine is negative, and sin(5π/3) = -sin(π/3) = -√3/2 ✓
General Solution Pattern
For equations of the form sin(Bx) = k, the general approach involves:
- Determining quadrants based on the sign of k
- Calculating principal solutions for Bx
- Finding the reference angle from |k|
- Adding multiples of 2π (the period of sine) to capture all solutions within the extended interval
This systematic approach ensures no solutions are missed when dealing with trigonometric functions that have coefficients affecting their period.
Conclusion
The equation sin(2x) = -√3/2 on the interval [0, 2π) yields four distinct solutions: x = 2π/3, 5π/6, 5π/3, 11π/6. This result emerges from understanding how the coefficient 2 affects the function's period, expanding our search interval from [0, 2π) to [0, 4π) for the argument 2x. Because of that, by methodically identifying the reference angle (π/3), locating the appropriate quadrants (III and IV), accounting for the extended interval through coterminal angles, and finally solving for x, we demonstrate a complete problem-solving framework applicable to similar trigonometric equations. This technique reinforces fundamental concepts of periodicity, reference angles, and the unit circle while showcasing the importance of scaling considerations in trigonometric problem-solving Simple as that..
Extending the Method: Solving More Complex Trigonometric Equations
The systematic approach illustrated for sin(2x) = −√3⁄2 does not stop at a single coefficient or constant. By adjusting the period‑scaling factor B and the target value k, we can solve a broad class of equations of the form
[ \sin(Bx)=k\qquad\text{or}\qquad\cos(Bx)=k, ]
where B ≠ 0 and k ∈ [−1, 1
