The point $(a, 2)$ is a stationary point on the graph $y = x^3 - x^2 - x + b$, where $a$ and $b$ are positive constants.

1. Find the derivative of the given curve equation

$y = x^3 - x^2 - x + b$
$\frac{dy}{dx} = 3x^2 - 2x - 1$

2. Set the derivative to zero to find the x-coordinates of the stationary points

At stationary points, $\frac{dy}{dx} = 0$:
$3x^2 - 2x - 1 = 0$
$(3x + 1)(x - 1) = 0$
$x = -\frac{1}{3}$ or $x = 1$

3. Identify the correct value for $a$ based on the given condition

Since $(a, 2)$ is a stationary point, $x = a$.
We are given that $a$ is a positive constant, so $a > 0$.
Therefore, $a = 1$ (shown).

1. Substitute the coordinates of the stationary point into the equation of the curve

Since $(a, 2)$ is on the curve and $a = 1$, substitute $x = 1$ and $y = 2$ into $y = x^3 - x^2 - x + b$:
$2 = (1)^3 - (1)^2 - (1) + b$

2. Solve for $b$

$2 = 1 - 1 - 1 + b$
$2 = -1 + b$
$b = 3$

1. Find the $x$-coordinate of the other stationary point by setting the derivative to zero

From earlier parts, $\frac{dy}{dx} = 3x^2 - 2x - 1$.
Setting $\frac{dy}{dx} = 0$:
$3x^2 - 2x - 1 = 0$
$(3x + 1)(x - 1) = 0$
$x = -\frac{1}{3}$ or $x = 1$
The other stationary point occurs at $x = -\frac{1}{3}$.

2. Find the corresponding $y$-coordinate

Substitute $x = -\frac{1}{3}$ and $b = 3$ into the equation of the curve:
$y = (-\frac{1}{3})^3 - (-\frac{1}{3})^2 - (-\frac{1}{3}) + 3$
$y = -\frac{1}{27} - \frac{1}{9} + \frac{1}{3} + 3$
$y = \frac{-1 - 3 + 9 + 81}{27}$
$y = \frac{86}{27}$

3. State the coordinates of the other stationary point

The coordinates are $(-\frac{1}{3}, \frac{86}{27})$.

1. Find the second derivative of the function

$\frac{dy}{dx} = 3x^2 - 2x - 1$
$\frac{d^2y}{dx^2} = 6x - 2$

2. Determine the nature of the stationary point at $x = 1$

When $x = 1$,
$\frac{d^2y}{dx^2} = 6(1) - 2 = 4 > 0$
Therefore, $(1, 2)$ is a minimum point.

3. Determine the nature of the stationary point at $x = -\frac{1}{3}$

When $x = -\frac{1}{3}$,
$\frac{d^2y}{dx^2} = 6\left(-\frac{1}{3}\right) - 2 = -4 < 0$
Therefore, the other stationary point is a maximum point.

1. Express the equation in terms of a single trigonometric function (cosine)

$2 \cos 2x - 3 \sec x + 2 = 0$
$2(2 \cos^2 x - 1) - \frac{3}{\cos x} + 2 = 0$

2. Simplify the equation into a polynomial in terms of $\cos x$

$4 \cos^2 x - 2 - \frac{3}{\cos x} + 2 = 0$
$4 \cos^2 x - \frac{3}{\cos x} = 0$
$4 \cos^3 x - 3 = 0$

3. Solve for $\cos x$

$\cos^3 x = \frac{3}{4}$
$\cos x = \sqrt[3]{\frac{3}{4}} \approx 0.90856$

4. Find the basic angle and solve for $x$ in the given domain

Basic angle $\alpha = \cos^{-1}(0.90856) \approx 0.431$ rad
Since $\cos x$ is positive, $x$ is in the 1st and 4th quadrants.
For $-\pi \le x \le \pi$:
$x = 0.431$ or $x = -0.431$