Q21: Range of $x$ when $\cot^{-1}(3x + 5) > \frac{\pi}{4}$

We write the right side as an inverse cotangent too, so that both sides can be compared directly.

$\cot\frac{\pi}{4} = 1$⁠, and $\frac{\pi}{4}$ lies in $(0, \pi)$⁠, the principal value range of $\cot^{-1}$⁠. So $\cot^{-1}1 = \frac{\pi}{4}$⁠.

As $x$ runs from $-\infty$ to $\infty$⁠, $\cot^{-1}x$ falls from $\pi$ to $0$⁠. Its derivative is negative everywhere:

For a strictly decreasing function, a bigger output comes from a smaller input:

Watch out: the inequality sign turns round. Keeping $>$ would give the wrong answer $x > -\frac43$⁠.

Subtract 5 from both sides, then divide by 3. Dividing by a positive number keeps the sign $<$⁠.

$\cot^{-1}$ is defined for every real number, so there is no other condition on $x$⁠.

Check with $x = -2$⁠: $3x + 5 = -1$⁠, and $\cot^{-1}(-1) = \frac{3\pi}{4}$⁠, which is greater than $\frac{\pi}{4}$⁠.

(The rule $\cot^{-1}(-u) = \pi - \cot^{-1}u$ is used on the second line.)