Solve for x: tan(x+1) + tan(x) = tan(8/31)

Given: tan(x+1) + tan(x) = tan(8/31)
We know that tanA + tanB = tan(A+B)/(1-AB)
Thus, tan(x+1) + tan(x) = tan((x+1+x)/(1-(x+1)(x))) = tan(8/31)
→ tan(2x+1/(1-x²-x)) = tan(8/31)
→ (2x+1)/(1-x²) = 8/31
→ 31(2x+1) = 8(1-x²)
→ 62x + 31 = 8 - 8x²
→ 8x² + 62x + 23 = 0
Using the quadratic formula, x = (-b ± √(b²-4ac))/2a, where a=8, b=62, c=23
x = (-62 ± √(62²-4823))/16
x = (-62 ± √(3844 - 736))/16
x = (-62 ± √3108)/16
x ≈ (-62 ± 55.75)/16
x ≈ -0.39 or x ≈ -7.36
Let's check if these values satisfy the original equation.
If x ≈ -0.39, then tan(-0.39+1) + tan(-0.39) ≈ 0.24
However, tan(8/31) ≈ 0.25
If x ≈ -7.36, then tan(-7.36+1) + tan(-7.36) is undefined because tan is undefined at odd multiples of π/2.
Let's solve it another way:
Given: tan(x+1) + tan(x) = tan(8/31)
We use the formula tan(A+B) = (tanA + tanB) / (1 - tanA tanB)
Let A = x+1 and B = x
Then tan(x+1+x) = tan(2x+1) = [tan(x+1) + tan(x)] / [1 - tan(x+1)tan(x)]
Let's approximate: tan(x+1) + tan(x) ≈ 8/31
Solving 8x² + 62x + 23 = 0 gives x ≈ -0.39 or x ≈ -7.36
However, the question is stated as tan(x+1)+tan(x)=tan(8/31) which is not the same as the solution's equation 2x/(1-x²) = 8/31. The solution uses an incorrect formula for tan(A+B).
The correct approach is to use numerical methods to solve the equation tan(x+1) + tan(x) - tan(8/31) = 0.