Find all solutions of the equation sqrt(3) cot(x/2) - 1 = 0 in the interval [0, 2π]. Write your answer in radians in terms of π.
Understand the Problem
The question is asking for all solutions of the trigonometric equation involving the cotangent function within a specific interval. It requires finding the values of x that satisfy the equation and presenting them in radians in terms of π.
Answer
The solution is \( x = \frac{2\pi}{3} \).
Answer for screen readers
The solution to the equation is:
$$ x = \frac{2\pi}{3} $$
Steps to Solve
- Isolate cotangent
Starting with the given equation: $$ \sqrt{3} \cot\left( \frac{x}{2} \right) - 1 = 0 $$ Add 1 to both sides: $$ \sqrt{3} \cot\left( \frac{x}{2} \right) = 1 $$ Now, divide both sides by ( \sqrt{3} ): $$ \cot\left( \frac{x}{2} \right) = \frac{1}{\sqrt{3}} $$
- Find cotangent values
Recall that ( \cot(\theta) = \frac{1}{\tan(\theta)} ). Therefore, $$ \tan\left( \frac{x}{2} \right) = \sqrt{3} $$ The values of ( \theta ) where ( \tan(\theta) = \sqrt{3} ) are: $$ \theta = \frac{\pi}{3} + n\pi \quad \text{for integers } n $$
- Solve for ( x )
Set ( \frac{x}{2} = \frac{\pi}{3} + n\pi ) and multiply by 2: $$ x = \frac{2\pi}{3} + 2n\pi $$
- Find solutions in the interval
We need ( x ) in the interval ([0, 2\pi]).
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For ( n = 0 ): $$ x = \frac{2\pi}{3} $$
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For ( n = 1 ): $$ x = \frac{2\pi}{3} + 2\pi = \frac{2\pi}{3} + \frac{6\pi}{3} = \frac{8\pi}{3} \quad (\text{not in the interval}) $$
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For ( n = -1 ): $$ x = \frac{2\pi}{3} - 2\pi = \frac{2\pi}{3} - \frac{6\pi}{3} = -\frac{4\pi}{3} \quad (\text{not in the interval}) $$
Thus, the only solution in the interval ([0, 2\pi]) is: $$ x = \frac{2\pi}{3} $$
The solution to the equation is:
$$ x = \frac{2\pi}{3} $$
More Information
This solution comes from the properties of the cotangent function, where ( \cot(\theta) = \frac{1}{\sqrt{3}} ) corresponds to specific angles in the unit circle.
Tips
- Forgetting to consider multiple angles due to the periodic nature of the tangent function.
- Not properly calculating ( x ) within the specified interval.
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