How do you solve the following equation $cot (\frac{x}{2}) = 1$ $cot (x/2)=1$ in the interval [0, 2pi]?

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How do you solve the following equation $cot (\frac{x}{2}) = 1$ $cot (x/2)=1$ in the interval [0, 2pi]?

1 Answer

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Answer 1 · 2016-09-11T12:00:26

$\frac{π}{2}$ $pi/2$

Explanation:

Trig table of special arcs and unit circle -->
$cot (\frac{x}{2}) = 1$ $cot (x/2) = 1$ --> $\frac{x}{2} = \frac{π}{4} + k π$ $x/2 = pi/4 + kpi$
General answers:
$x = \frac{π}{2} + k 2 π$ $x = pi/2 + k2pi$
Answer in interval (0, 2pi) --> k = 0
$x = \frac{π}{2}$ $x = pi/2$

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How do you solve the following equation $cot (\frac{x}{2}) = 1$ $cot (x/2)=1$ in the interval [0, 2pi]?

1 Answer

Explanation:

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How do you solve the following equation cot (x/2)=1cot(x2)=1cot (x/2)=1 in the interval [0, 2pi]?

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Explanation:

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How do you solve the following equation $cot (\frac{x}{2}) = 1$ $cot (x/2)=1$ in the interval [0, 2pi]?