cot-1[tan(2x)] + cot-1[tan(3x)] = x
Since tangent and cotangent have period 180°,
cot-1[tan(2x±180°k)] + cot-1[tan(3x±180°)] = x
replace tan(2x±180°k) by cot[90°-(2x±180°k)] and tan(3x±180°k) by
cot[90°-(3x+180°k)]
cot-1{cot[90°-(2x±180°)]} + cot-1{cot[90°-(3x±180°)]} = x
[90°-(2x±180°k)] + [90°-(3x±180°k)] = x
90° - 2x ∓ 180°k + 90° - 3x ∓ 180°k = x
180° ∓ 360°k - 5x = x
180° ∓ 360°k = 6x
180°(1 ∓ 2k) = 6x
30°(1 ∓ 2k) = x
Since k can take on negative integer values, we
can just use a + sign.
30°(2k + 1) = x
So x can be any odd multiple of 30°.
However we must discard this because the original equation contains
tan(3x), which would be undefined, for 3x would be odd multiples of 90°
and tangents of odd multiples of 90° is undefined.
So after all that, we find that there is no solution.
Edwin
