Question 70665

Concept

An inverse sine expression sin−1t is defined only when its argument satisfies −1 ≤ t ≤ 1, while a real square root √A exists only when A ≥ 0. Whenever an equation mixes square roots with sin−1, intersecting these two domain conditions often fixes the admissible values before any algebra is done. The companion identity tan−1a + cot−1a = π/2 for a ≥ 0, together with sin(cot−1a) = 1/√(1 + a2), turns a sum of two inverse functions into a single equation.

Application

1. Put u = x2 + x, so the equation reads tan−1√u + sin−1√(u + 1) = π/2.

2. The term √u is real only when u ≥ 0.

3. The term sin−1√(u + 1) requires 0 ≤ √(u + 1) ≤ 1, that is 0 ≤ u + 1 ≤ 1, that is −1 ≤ u ≤ 0.

4. Intersecting u ≥ 0 with −1 ≤ u ≤ 0 leaves exactly one admissible value, u = 0.

5. Check u = 0 in the equation: tan−10 + sin−11 = 0 + π/2 = π/2, so u = 0 really does satisfy it.

6. Solve u = 0 for x: x2 + x = 0, so x(x + 1) = 0, giving x = 0 or x = −1.

Cross-check

The same value of u follows algebraically, without using the domain argument. Write a = √u ≥ 0. Since π/2 − tan−1a = cot−1a, the equation becomes sin−1√(a2 + 1) = cot−1a. Taking the sine of both sides and using sin(cot−1a) = 1/√(1 + a2) gives √(a2 + 1) = 1/√(1 + a2), hence (1 + a2)2 = 1, so a2 = 0 and u = 0.

A value outside this set shows the mechanism clearly: at x = 1 the radicand becomes x2 + x + 1 = 3, and sin−1√3 is undefined because √3 > 1, so 1 cannot belong to the solution set.

Result

The equation is satisfied exactly by x = 0 and x = −1.