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What is the general solution of the equation: `tan^2 theta + 2sqrt3 tan theta = 1?` |
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Answer» Correct Answer - `theta=n pi + pi/12, n in Z or theta = mpi - (5pi)/12; m in Z` `tan^(2) theta + 2 sqrt3 tan theta = 1` or `(tan theta + sqrt3)^(2) = 4` or `tan theta = 2 - sqrt3 or -2 - sqrt3` or `tan theta = tan 15^(@), - tan 75^(@)` `= "tan"(pi)/(12), tan (-(5pi)/(12))` From `tan theta = "tan" (pi)/(12)`, we get `theta = n pi + (pi)/(12), n in Z` From `tan theta = tan (-(5pi)/(12))`, we get `theta = m pi - (5pi)/(12), m in Z` |
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