Trigonometry — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for trigonometry.

UT-1: Ambiguous Case of the Sine Rule

Question:

In triangle $ABC$, $a = 8$, $b = 10$, and $A = 35°$.

(a) Find the two possible values of angle $B$, giving your answers to 1 decimal place.

(b) For each value of $B$, find the corresponding value of angle $C$ and the length of side $c$.

(c) Explain why a student who applies the cosine rule to find $B$ directly (without first finding $c$) would fail to discover the ambiguous case.

[Difficulty: hard. Tests the ambiguous case of the sine rule where two valid triangles exist, and why alternative methods miss the second solution.]

Solution:

(a) By the sine rule:

$\frac◆LB◆\sin B◆RB◆◆LB◆b◆RB◆ = \frac◆LB◆\sin A◆RB◆◆LB◆a◆RB◆$

$\sin B = \frac◆LB◆b \sin A◆RB◆◆LB◆a◆RB◆ = \frac◆LB◆10 \sin 35°◆RB◆◆LB◆8◆RB◆ = \frac◆LB◆10 \times 0.5736◆RB◆◆LB◆8◆RB◆ = 0.7170$

Since $\sin B = 0.7170$ and $B$ is an angle in a triangle ($0° \lt B \lt 180°$), there are two solutions:

$B_1 = \arcsin(0.7170) = 45.8°$

$B_2 = 180° - 45.8° = 134.2°$

Both are valid: $B_1 + A = 80.8° \lt 180°$ and $B_2 + A = 169.2° \lt 180°$.

(b) For $B_1 = 45.8°$:

$C_1 = 180° - 35° - 45.8° = 99.2°$

$c_1 = \frac◆LB◆a \sin C_1◆RB◆◆LB◆\sin A◆RB◆ = \frac◆LB◆8 \sin 99.2°◆RB◆◆LB◆\sin 35°◆RB◆ = \frac◆LB◆8 \times 0.9863◆RB◆◆LB◆0.5736◆RB◆ = 13.76$

For $B_2 = 134.2°$:

$C_2 = 180° - 35° - 134.2° = 10.8°$

$c_2 = \frac◆LB◆8 \sin 10.8°◆RB◆◆LB◆\sin 35°◆RB◆ = \frac◆LB◆8 \times 0.1874◆RB◆◆LB◆0.5736◆RB◆ = 2.61$

(c) The cosine rule for finding $B$ is $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$. This requires knowing $c$, which is not given. The cosine rule is not directly applicable with the SSA (two sides and a non-included angle) configuration.

More fundamentally, $\cos$ is injective on $(0°, 180°)$ (it is strictly decreasing), so the cosine rule can only ever yield one value for $B$. The ambiguity is an inherent property of the sine rule, because $\sin$ is symmetric about $90°$ on $(0°, 180°)$: $\sin\theta = \sin(180° - \theta)$.

---

UT-2: Cancelling $\sin\theta$ When $\sin\theta = 0$

Question:

(a) Solve the equation $\sin\theta = 2\sin\theta\cos\theta$ for $\theta \in [0°, 360°)$.

(b) A student's working is shown below. Identify the error, state which solutions are lost, and explain why.

$\sin\theta = 2\sin\theta\cos\theta$

Dividing both sides by $\sin\theta$:

$1 = 2\cos\theta$

$\cos\theta = \frac{1}{2}$

$\theta = 60°$ or $\theta = 300°$

(c) Find all solutions of $\tan 2x = \tan x$ for $x \in [0, \pi)$, taking care not to lose solutions.

[Difficulty: hard. Tests the common error of dividing by a trigonometric expression that may equal zero, which systematically discards valid solutions.]

Solution:

(a) $\sin\theta = 2\sin\theta\cos\theta$

$\sin\theta - 2\sin\theta\cos\theta = 0$

$\sin\theta(1 - 2\cos\theta) = 0$

Case 1: $\sin\theta = 0 \implies \theta = 0°, 180°, 360°$

Case 2: $1 - 2\cos\theta = 0 \implies \cos\theta = \frac{1}{2} \implies \theta = 60°, 300°$

All solutions: $\theta = 0°, 60°, 180°, 300°, 360°$.

(b) The error is dividing both sides by $\sin\theta$. This is only valid when $\sin\theta \neq 0$. By dividing, the student implicitly assumes $\sin\theta \neq 0$, which discards the solutions $\theta = 0°, 180°, 360°$. The correct approach is to factorise, not to divide.

(c) $\tan 2x = \tan x$ for $x \in [0, \pi)$.

First, establish the domain. Both $\tan x$ and $\tan 2x$ must be defined:

• $\tan x$ undefined at $x = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$
• $\tan 2x$ undefined at $2x = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$ and $2x = \frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆$, i.e. $x = \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$ and $x = \frac◆LB◆3\pi◆RB◆◆LB◆4◆RB◆$

So the domain excludes $x = \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆, \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆, \frac◆LB◆3\pi◆RB◆◆LB◆4◆RB◆$.

Using the double angle formula $\tan 2x = \frac◆LB◆2\tan x◆RB◆◆LB◆1 - \tan^2 x◆RB◆$:

$\frac◆LB◆2\tan x◆RB◆◆LB◆1 - \tan^2 x◆RB◆ = \tan x$

Do NOT divide by $\tan x$. Bring everything to one side:

$\frac◆LB◆2\tan x - \tan x(1 - \tan^2 x)◆RB◆◆LB◆1 - \tan^2 x◆RB◆ = 0$

$\frac◆LB◆\tan x + \tan^3 x◆RB◆◆LB◆1 - \tan^2 x◆RB◆ = 0$

$\frac◆LB◆\tan x(1 + \tan^2 x)◆RB◆◆LB◆1 - \tan^2 x◆RB◆ = 0$

Since $1 + \tan^2 x = \sec^2 x \geq 1$ for all $x$ where $\tan x$ is defined, the numerator is zero when $\tan x = 0$:

$\tan x = 0 \implies x = 0$

(The solution $x = \pi$ is excluded by the interval $[0, \pi)$.)

Verify: $\tan 0 = 0$ and $\tan 0 = 0$. Confirmed.

Solution: $x = 0$ only.

---

UT-3: Solving $\cos(3x) = \frac{1}{2}$ with Full Periodicity

Question:

(a) Solve $\cos(3x) = \frac{1}{2}$ for $x \in [0, 2\pi)$. A common error produces only 2 or 4 solutions. Find all solutions.

(b) A student argues: "$\cos(3x) = \frac{1}{2}$ means $3x = \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆$ or $3x = \frac◆LB◆5\pi◆RB◆◆LB◆3◆RB◆$, giving $x = \frac◆LB◆\pi◆RB◆◆LB◆9◆RB◆$ or $x = \frac◆LB◆5\pi◆RB◆◆LB◆9◆RB◆$." Explain the error in this reasoning and state how many solutions the student is missing.

(c) The curve $y = \cos(3x)$ intersects the line $y = \frac{1}{2}$ at $N$ distinct points in the interval $[0, 2\pi)$. Find $N$ and the sum of all $x$-coordinates of the intersection points.

[Difficulty: hard. Tests understanding that multiplying the argument of a trigonometric function by a constant changes the period, requiring systematic enumeration of all solutions within the given interval.]

Solution:

(a) $\cos(3x) = \frac{1}{2}$.

Let $\theta = 3x$. Since $x \in [0, 2\pi)$, we have $\theta \in [0, 6\pi)$.

$\cos\theta = \frac{1}{2}$ gives $\theta = \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆ + 2n\pi$ or $\theta = \frac◆LB◆5\pi◆RB◆◆LB◆3◆RB◆ + 2n\pi$ for $n \in \mathbb{Z}$.

Systematically listing all $\theta \in [0, 6\pi)$:

$n$	$\theta = \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆ + 2n\pi$	$\theta = \frac◆LB◆5\pi◆RB◆◆LB◆3◆RB◆ + 2n\pi$
0	$\frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆$	$\frac◆LB◆5\pi◆RB◆◆LB◆3◆RB◆$
1	$\frac◆LB◆7\pi◆RB◆◆LB◆3◆RB◆$	$\frac◆LB◆11\pi◆RB◆◆LB◆3◆RB◆$
2	$\frac◆LB◆13\pi◆RB◆◆LB◆3◆RB◆$	$\frac◆LB◆17\pi◆RB◆◆LB◆3◆RB◆$

All six values are in $[0, 6\pi)$ since $\frac◆LB◆17\pi◆RB◆◆LB◆3◆RB◆ = 5\frac{2}{3}\pi \lt 6\pi$.

Dividing by 3:

$x = \frac◆LB◆\pi◆RB◆◆LB◆9◆RB◆,\quad \frac◆LB◆5\pi◆RB◆◆LB◆9◆RB◆,\quad \frac◆LB◆7\pi◆RB◆◆LB◆9◆RB◆,\quad \frac◆LB◆11\pi◆RB◆◆LB◆9◆RB◆,\quad \frac◆LB◆13\pi◆RB◆◆LB◆9◆RB◆,\quad \frac◆LB◆17\pi◆RB◆◆LB◆9◆RB◆$

(b) The student correctly identifies the principal solutions for $3x$ but fails to account for the fact that the period of $\cos(3x)$ is $\frac◆LB◆2\pi◆RB◆◆LB◆3◆RB◆$, not $2\pi$. When $x$ ranges over $[0, 2\pi)$, the argument $3x$ ranges over $[0, 6\pi)$, which spans three full periods of cosine. The student finds solutions from only the first period, missing the four solutions from the second and third periods. The student is missing 4 out of 6 solutions.

(c) $N = 6$ (from part (a)).

The sum of all $x$-coordinates:

$S = \frac◆LB◆\pi◆RB◆◆LB◆9◆RB◆ + \frac◆LB◆5\pi◆RB◆◆LB◆9◆RB◆ + \frac◆LB◆7\pi◆RB◆◆LB◆9◆RB◆ + \frac◆LB◆11\pi◆RB◆◆LB◆9◆RB◆ + \frac◆LB◆13\pi◆RB◆◆LB◆9◆RB◆ + \frac◆LB◆17\pi◆RB◆◆LB◆9◆RB◆ = \frac◆LB◆60\pi◆RB◆◆LB◆9◆RB◆ = \frac◆LB◆20\pi◆RB◆◆LB◆3◆RB◆$

---

Integration Tests

Tests synthesis of trigonometry with other topics. Requires combining concepts from multiple units.

IT-1: Related Rates with Trigonometric Functions (with Differentiation)

Question:

A ladder of length 5 metres is leaning against a vertical wall. The foot of the ladder is pulled away from the wall horizontally at a constant rate of $0.5$ m/s.

(a) Find the rate at which the top of the ladder is sliding down the wall when the foot of the ladder is 3 metres from the wall.

(b) Find the angle $\theta$ between the ladder and the ground at the instant when the top of the ladder is sliding down at the same speed as the foot is being pulled away.

(c) Show that $\theta$ is always decreasing, and find the rate of change of $\theta$ in rad/s when $\theta = \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$.

[Difficulty: hard. Combines implicit differentiation with trigonometric relationships in a kinematics context.]

Solution:

(a) Let $x$ be the distance from the foot of the ladder to the wall, and $y$ be the height of the top of the ladder above the ground.

By Pythagoras' theorem: $x^2 + y^2 = 25$.

Differentiating implicitly with respect to $t$:

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies x\frac{dx}{dt} + y\frac{dy}{dt} = 0$

Given $\frac{dx}{dt} = 0.5$. When $x = 3$: $y = \sqrt{25 - 9} = 4$.

$3(0.5) + 4\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{3}{8} \text{ m/s}$

The negative sign confirms the top is sliding down.

(b) We need $\left\lvert\frac{dy}{dt}\right\rvert = \left\lvert\frac{dx}{dt}\right\rvert = 0.5$.

From $x\frac{dx}{dt} + y\frac{dy}{dt} = 0$: $\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}$.

$\left\lvert-\frac{x}{y} \cdot 0.5\right\rvert = 0.5 \implies \frac{x}{y} = 1 \implies x = y$

Since $x^2 + y^2 = 25$ and $x = y$: $2x^2 = 25$, so $x = \frac◆LB◆5◆RB◆◆LB◆\sqrt{2}◆RB◆$.

$\cos\theta = \frac{x}{5} = \frac◆LB◆1◆RB◆◆LB◆\sqrt{2}◆RB◆$, giving $\theta = \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$.

(c) $\cos\theta = \frac{x}{5}$.

Differentiating with respect to $t$:

$-\sin\theta \cdot \frac◆LB◆d\theta◆RB◆◆LB◆dt◆RB◆ = \frac{1}{5}\frac{dx}{dt}$

$\frac◆LB◆d\theta◆RB◆◆LB◆dt◆RB◆ = -\frac◆LB◆1◆RB◆◆LB◆5\sin\theta◆RB◆ \cdot \frac{dx}{dt} = -\frac◆LB◆0.5◆RB◆◆LB◆5\sin\theta◆RB◆ = -\frac◆LB◆1◆RB◆◆LB◆10\sin\theta◆RB◆$

Since $\sin\theta \gt 0$ for $0 \lt \theta \lt \pi$, we have $\frac◆LB◆d\theta◆RB◆◆LB◆dt◆RB◆ \lt 0$, confirming $\theta$ is always decreasing.

When $\theta = \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$: $\sin\frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆\sqrt{2}◆RB◆$.

$\frac◆LB◆d\theta◆RB◆◆LB◆dt◆RB◆ = -\frac◆LB◆1◆RB◆◆LB◆10 \cdot \frac◆LB◆1◆RB◆◆LB◆\sqrt{2}◆RB◆◆RB◆ = -\frac◆LB◆\sqrt{2}◆RB◆◆LB◆10◆RB◆ \text{ rad/s}$

---

IT-2: Area Enclosed by a Parametric Trigonometric Curve (with Coordinate Geometry)

Question:

A curve is defined parametrically by:

$x = a(\cos t + \cos 2t), \quad y = a(\sin t + \sin 2t)$

for $0 \leq t \leq 2\pi$, where $a \gt 0$.

(a) Show that the curve is closed.

(b) Find the total area enclosed by the curve, giving your answer in terms of $a$.

(c) Find the coordinates of all points on the curve where the tangent is horizontal, giving exact answers in terms of $a$.

[Difficulty: hard. Combines parametric differentiation, trigonometric identities, and definite integration for area under a parametric curve.]

Solution:

(a) At $t = 0$: $x = a(1 + 1) = 2a$, $y = a(0 + 0) = 0$.

At $t = 2\pi$: $x = a(\cos 2\pi + \cos 4\pi) = a(1+1) = 2a$, $y = a(\sin 2\pi + \sin 4\pi) = 0$.

The curve starts and ends at $(2a, 0)$, so it is closed.

(b) The area enclosed by a parametric curve $(x(t), y(t))$ traced once anticlockwise is:

$A = \left\lvert\int_0^{2\pi} y\frac{dx}{dt}\, dt\right\rvert$

$\frac{dx}{dt} = a(-\sin t - 2\sin 2t)$

$y\frac{dx}{dt} = a(\sin t + \sin 2t) \cdot a(-\sin t - 2\sin 2t) = -a^2(\sin t + \sin 2t)(\sin t + 2\sin 2t)$

Expanding:

$= -a^2[\sin^2 t + 3\sin t\sin 2t + 2\sin^2 2t]$

Integrate each term separately over $[0, 2\pi]$:

$\int_0^{2\pi}\sin^2 t\, dt = \int_0^{2\pi}\frac◆LB◆1 - \cos 2t◆RB◆◆LB◆2◆RB◆\, dt = \left[\frac{t}{2} - \frac◆LB◆\sin 2t◆RB◆◆LB◆4◆RB◆\right]_0^{2\pi} = \pi$

$\int_0^{2\pi}\sin t\sin 2t\, dt = \int_0^{2\pi}2\sin^2 t\cos t\, dt = 2\left[\frac◆LB◆\sin^3 t◆RB◆◆LB◆3◆RB◆\right]_0^{2\pi} = 0$

$\int_0^{2\pi}\sin^2 2t\, dt = \int_0^{2\pi}\frac◆LB◆1 - \cos 4t◆RB◆◆LB◆2◆RB◆\, dt = \left[\frac{t}{2} - \frac◆LB◆\sin 4t◆RB◆◆LB◆8◆RB◆\right]_0^{2\pi} = \pi$

Therefore:

$\int_0^{2\pi} y\frac{dx}{dt}\, dt = -a^2[\pi + 3(0) + 2\pi] = -3\pi a^2$

$A = \lvert -3\pi a^2 \rvert = 3\pi a^2$

(c) The tangent is horizontal when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$.

$\frac{dy}{dt} = a(\cos t + 2\cos 2t) = 0$

$\cos t + 2\cos 2t = 0$

Using $\cos 2t = 2\cos^2 t - 1$:

$\cos t + 2(2\cos^2 t - 1) = 0 \implies 4\cos^2 t + \cos t - 2 = 0$

Let $u = \cos t$:

$u = \frac◆LB◆-1 \pm \sqrt{1 + 32}◆RB◆◆LB◆8◆RB◆ = \frac◆LB◆-1 \pm \sqrt{33}◆RB◆◆LB◆8◆RB◆$

Both roots lie in $[-1, 1]$: $u_1 = \frac◆LB◆\sqrt{33}-1◆RB◆◆LB◆8◆RB◆ \approx 0.593$ and $u_2 = -\frac◆LB◆\sqrt{33}+1◆RB◆◆LB◆8◆RB◆ \approx -0.843$.

Each gives two values of $t$ in $[0, 2\pi)$, producing four horizontal tangent points.

For $u_1 = \frac◆LB◆\sqrt{33}-1◆RB◆◆LB◆8◆RB◆$: $t = \pm\arccos\left(\frac◆LB◆\sqrt{33}-1◆RB◆◆LB◆8◆RB◆\right)$. Substituting into $x$ and $y$:

$x = a\left(\frac◆LB◆\sqrt{33}-1◆RB◆◆LB◆8◆RB◆\right) + a\left(2\left(\frac◆LB◆\sqrt{33}-1◆RB◆◆LB◆8◆RB◆\right)^2 - 1\right)$

$= a\left(\frac◆LB◆\sqrt{33}-1◆RB◆◆LB◆8◆RB◆ + \frac◆LB◆34-2\sqrt{33}◆RB◆◆LB◆32◆RB◆ - 1\right) = a \cdot \frac◆LB◆4\sqrt{33}-4+34-2\sqrt{33}-32◆RB◆◆LB◆32◆RB◆ = \frac◆LB◆a(\sqrt{33}-1)◆RB◆◆LB◆16◆RB◆$

By symmetry, the four horizontal tangent points are:

$\left(\frac◆LB◆a(\sqrt{33}-1)◆RB◆◆LB◆16◆RB◆,\quad \pm\frac{a}{16}\sqrt◆LB◆30+2\sqrt{33}◆RB◆\right)$

$\left(\frac◆LB◆-a(\sqrt{33}+1)◆RB◆◆LB◆16◆RB◆,\quad \pm\frac{a}{16}\sqrt◆LB◆30-2\sqrt{33}◆RB◆\right)$

---

IT-3: Inverse Trigonometric Composition and Domain Restrictions (with Functions)

Question:

The function $f$ is defined by $f(x) = \cos x$ on the domain $\left[0, \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right]$.

(a) State the range of $f$ and explain why $f^{-1}$ exists on this domain.

(b) Let $g(x) = \arccos x$ (principal inverse cosine, domain $[-1, 1]$, range $[0, \pi]$). Find $g(f(x))$ and $f(g(x))$, stating the domain of each composition.

(c) A student claims $f^{-1}(f(\pi)) = \pi$. Explain why this is incorrect.

(d) The function $h$ is defined by $h(x) = \arcsin(\cos x)$ for $x \in \mathbb{R}$. Express $h(x)$ as a piecewise function and find all $x$ for which $h(x) = x$.

[Difficulty: hard. Tests the interplay between trigonometric functions and their inverses, requiring careful attention to domain and range restrictions.]

Solution:

(a) On $\left[0, \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right]$, $f'(x) = -\sin x \lt 0$ for $0 \lt x \lt \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$, so $f$ is strictly decreasing and therefore injective.

$f(0) = 1$ and $f\!\left(\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right) = 0$, so the range is $[0, 1]$.

Since $f$ is injective, $f^{-1}$ exists. Its domain is $[0, 1]$ and its range is $\left[0, \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right]$.

(b) $g(f(x)) = \arccos(\cos x)$: defined when $\cos x \in [-1, 1]$, which is always true. Domain: $\left[0, \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right]$.

For $x \in \left[0, \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right]$: since $x \in [0, \pi]$ (the range of $\arccos$), we have $\arccos(\cos x) = x$. So $g(f(x)) = x$.

$f(g(x)) = \cos(\arccos x)$: defined when $\arccos x \in \left[0, \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right]$, i.e. $x \in [0, 1]$.

For $x \in [0, 1]$: $\cos(\arccos x) = x$. So $f(g(x)) = x$.

The domains differ: $g \circ f$ has domain $\left[0, \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right]$ while $f \circ g$ has domain $[0, 1]$.

(c) $\pi$ is not in the domain of $f$ (which is $\left[0, \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right]$), so $f(\pi)$ is undefined. The expression $f^{-1}(f(\pi))$ is therefore meaningless.

Even if $f$ were extended to $\mathbb{R}$, then $f(\pi) = -1$, but $-1$ is not in the domain of $f^{-1}$ (which is $[0, 1]$, the range of $f$ on its original domain). So $f^{-1}(-1)$ would also be undefined.

(d) $h(x) = \arcsin(\cos x)$. Using $\cos x = \sin\!\left(\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ - x\right)$:

$h(x) = \arcsin\!\left(\sin\!\left(\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ - x\right)\right)$

Since $\arcsin(\sin\theta) = \theta$ only when $\theta \in \left[-\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆, \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆\right]$:

When $-\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ \leq \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ - x \leq \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$, i.e. $0 \leq x \leq \pi$: $h(x) = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ - x$.

When $\pi \leq x \leq 2\pi$: $\cos x = \cos(2\pi - x)$ and $2\pi - x \in [0, \pi]$, so $h(x) = \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ - (2\pi - x) = x - \frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆$.

By periodicity (period $2\pi$):

$h(x) = \begin{cases} \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ - x + 2n\pi & \text{if } 2n\pi \leq x \leq (2n+1)\pi \\ x - \frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆ + 2n\pi & \text{if } (2n+1)\pi \leq x \leq (2n+2)\pi \end{cases}$

for $n \in \mathbb{Z}$.

To find $h(x) = x$: on $[0, \pi]$, $\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆ - x = x \implies x = \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆$.

On $[\pi, 2\pi]$: $x - \frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆ = x \implies 0 = \frac◆LB◆3\pi◆RB◆◆LB◆2◆RB◆$, impossible.

By periodicity: $x = \frac◆LB◆\pi◆RB◆◆LB◆4◆RB◆ + 2n\pi$ for $n \in \mathbb{Z}$.