Let X be the solution set of the equation Ax = I, where A=01−14−343−34 and I is the corresponding unit matrix and x⊆N, then the minimum value of ∑cosx⁡θ+sinx⁡θ, θ∈(0,π/2), is ___________.

A=01−14−343−34⇒ A2=A⋅A=01−14−343−34⋅01−14−343−34=0+4−30−3+30+4−40−12+124+9−12−4−12+160−12+123+9−12−3−12+16=100010001⇒ A2=I⇒A4=A6=A8=INow Ax = I⇒ x=2, 4, 6, 8, …∴     Σcosx⁡θ+sinx⁡θ    =cos2⁡θ+sin2⁡θ+cos4⁡θ+sin4⁡θ+cos6⁡θ+sin6⁡θ+...    =cos2⁡θ+cos4⁡θ+cos6⁡θ+....+sin4⁡θ+sin4⁡θ+sin6⁡θ....    =cos2⁡θ1−cos2⁡θ+sin2⁡θ1−sin2⁡θ    =cot2⁡θ+tan2⁡θwhich has minimum value 2