How do you solve $tan(theta)= -2.5$ over the interval 0 to 2pi?

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Answer 1 · 2016-10-03T07:41:16

$111.8^o and 291.8^o$ , nearly.

The calculator gives

$theta=tan^(-1)(-2.5)= -68.19859...^o=-68.2^o$ , nearly.

It is programmed to give only

the principal value that $in [-90^o, 90^o]$ ...

The general value is given by $theta = kpi$ +principal ( or any ) value,

$k = 0, +-1, +-2, +-3, ...$ .

Here, $theta = (180k-68.2)^o, k = 0, +-1, +-2, +-3, ...$

Upon setting k = 1 and 2, $theta =111.8^o and 291.8^o in [0. 360^o]$

The calculator gives tangents of these angles as $-2.500....$ ..

How do you solve tan(theta)= -2.5tan(theta)= -2.5 over the interval 0 to 2pi?