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To find the values of $ \tan(\theta) $ that satisfy the equation $ \tan(\theta) = 2 $ in the interval $ [0, 2\pi] $, we need to determine the angles where the tangent function equals 2.

First, recall that the tangent function is periodic with period $ \pi $, and the angles where $ \tan(\theta) = 2 $ are:

$ \theta_1 = \arctan(2) $

and

$ \theta_2 = \arctan(2) + \pi $

Because the tangent function repeats every $ \pi $ radians, we only need to check within one period:

$ \theta_1 = \arctan(2) $

$ \theta_2 = \arctan(2) + \pi $

Thus, the solutions within $ [0, 2\pi] $ are:

$ \theta = \arctan(2) $

and

$ \theta = \arctan(2) + \pi $

To find the values of $ an( heta) $ that satisfy the equation $ an( heta) = 2 $ in the interval $ [0, 2pi] $, we need to determine the angles where the tangent function equals 2.

The tangent function is periodic with period $ pi $, so we only need to find solutions within one period:

$ heta_1 = arctan(2) $

and

$ heta_2 = arctan(2) + pi $

Since these angles lie within $ [0, 2pi] $, the solutions are:

$ heta = arctan(2) $

and

$ heta = arctan(2) + pi $

Find the values of $ an( heta) $ that satisfy $ an( heta) = 2 $ within $ [0, 2pi] $.

Solutions:

$ heta = arctan(2) $

and

$ heta = arctan(2) + pi $

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