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....#..