Mathematics Advanced • Year 11 • Module 2 • Lesson 13
Solving Trigonometric Equations Graphically
Practise HSC-style writing on solving with quadrants, counting in extended domains, and reasoning about transformed-equation solutions.
1. Short-answer questions
1.1 By sketching $y = \sin x$ and $y = -\dfrac{1}{2}$ on the same axes, solve $\sin x = -\dfrac{1}{2}$ for $x \in [0, 2\pi]$. Give exact values. 3 marks Band 3
1.2 Consider $2\cos x + 1 = 0$.
(a) Rewrite in the form $\cos x = k$.
(b) Use a sketch of $y = \cos x$ and $y = k$ to find all solutions in $[0, 2\pi]$.
(c) State how many solutions the same equation has in $[-2\pi, 2\pi]$. 4 marks Band 3-4
1.3 By sketching the graphs of $y = \sin x$ and $y = \cos x$ on the same axes, solve $\sin x = \cos x$ in $[0, 2\pi]$. Justify both solutions using the graph, then verify each algebraically. 4 marks Band 4
Stuck on 1.3? At an intersection both functions give the same $y$-value; divide by $\cos x$ to convert to $\tan x = 1$.2. Extended response
2.1 A music synthesiser produces a tone whose amplitude $A(t)$ (in arbitrary units) is $A(t) = 3\sin(\pi t) + 1$, where $t$ is in seconds. The trigger circuit on a downstream device fires every time $A(t) = 2.5$.
(a) Sketch $y = A(t)$ for $t \in [0, 4]$ and superimpose $y = 2.5$. Identify the number of solutions of $A(t) = 2.5$ in this domain using a period-count argument.
(b) Find all exact times $t$ in $[0, 2]$ when the trigger fires. Show the algebra explicitly (rearrange to $\sin(\pi t) = \frac{1}{2}$, then use quadrant reasoning).
(c) A student claims the trigger should fire every $1$ second because "the period of $\sin(\pi t)$ is $2$ s and there are 2 firings per period, so $2/2 = 1$ s gap". Evaluate this claim. Compute the actual gaps between consecutive firings (use the exact times from part b) and explain why the spacing is uneven — connecting your answer to one of the three lesson Traps. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — identifies period of $A$ is $2$ s, so $[0, 4]$ contains 2 full periods.
• 1 mark — concludes there are $2 \times 2 = 4$ solutions in $[0, 4]$.
Part (b) — 4 marks
• 1 mark — rearranges to $\sin(\pi t) = 1/2$.
• 1 mark — identifies reference angle $\pi/6$ in Q1.
• 1 mark — uses Q1 and Q2 (since $\sin > 0$): $\pi t = \pi/6$ and $\pi t = \pi - \pi/6 = 5\pi/6$.
• 1 mark — solves: $t = 1/6$ s and $t = 5/6$ s.
Part (c) — 2 marks
• 1 mark — computes gaps: $5/6 - 1/6 = 4/6 = 2/3$ s, then $1/6 + 2 - 5/6 = 4/3$ s — uneven.
• 1 mark — explains the "averaged spacing $1$ s" is wrong because the two solutions per period are positioned asymmetrically about the peak (Trap 01 — only counting one solution per cycle, or assuming uniform spacing).
Your response:
Stuck on (c)? The two solutions per period are symmetric about the peak ($\pi/2$ inside the bracket); they're not evenly spaced unless they happen to land on the midline.How did this worksheet feel?
What I'll revisit before next class:
1.1 — $\sin x = -1/2$ (3 marks)
Sample response. Sine is negative in Q3 and Q4. Reference angle: $\sin(\pi/6) = 1/2$, so $\alpha = \pi/6$. Q3: $\pi + \pi/6 = 7\pi/6$. Q4: $2\pi - \pi/6 = 11\pi/6$. $x = 7\pi/6, 11\pi/6$.
Marking notes. 1 mark — correct quadrants identified. 1 mark — reference angle. 1 mark — both solutions correct. Common error: forgetting the Q4 solution (Trap 01 — only finding one solution per cycle).
1.2 — $2\cos x + 1 = 0$ (4 marks)
Sample response.
(a) $\cos x = -1/2$ [1].
(b) Cosine negative in Q2 and Q3. Reference angle $\pi/3$. Q2: $\pi - \pi/3 = 2\pi/3$. Q3: $\pi + \pi/3 = 4\pi/3$. $x = $ $2\pi/3, 4\pi/3$ [2].
(c) Domain $[-2\pi, 2\pi]$ spans $4\pi$ = 2 periods, with 2 solutions per period, so $4$ solutions [1].
Marking notes. (a) 1 mark. (b) 1 for quadrant reasoning, 1 for both solutions. (c) 1 mark for period count argument.
1.3 — $\sin x = \cos x$ (4 marks)
Sample response. Sketch $y = \sin x$ and $y = \cos x$. They intersect where both functions have the same $y$-value: first intersection in Q1 just past $\pi/4$ at $y = \sqrt{2}/2$, then again in Q3 at $5\pi/4$ at $y = -\sqrt{2}/2$ [2]. Algebraic verification: divide both sides by $\cos x$ (valid since $\cos x \neq 0$ at the solutions): $\tan x = 1 \Rightarrow x = \pi/4 + n\pi$. In $[0, 2\pi]$: $x = $ $\pi/4, 5\pi/4$ [2].
Marking notes. 2 marks for the sketch + identifying the two intersections. 2 marks for the algebraic verification via $\tan x = 1$. Common error: students give only the Q1 solution because they "see" only one obvious intersection on the sketch.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Number of solutions. The period of $A(t) = 3\sin(\pi t) + 1$ is $T = 2\pi/\pi = 2$ seconds, so $[0, 4]$ contains $4/2 = 2$ complete periods. [1 mark] Since $2.5$ lies between the midline ($1$) and the maximum ($4$), the horizontal line $y = 2.5$ cuts each cycle in 2 points (one rising, one falling). Total: $2 \times 2 = 4$ solutions in $[0, 4]$. [1 mark]
Part (b) — Exact times in $[0, 2]$. Rearrange:
$3\sin(\pi t) + 1 = 2.5 \Rightarrow 3\sin(\pi t) = 1.5 \Rightarrow \sin(\pi t) = 1/2$. [1 mark]
Reference angle in Q1: $\sin(\pi/6) = 1/2$, so $\pi t = \pi/6$ gives $t = 1/6$ s (rising crossing). [1 mark]
Sine is also positive in Q2: $\pi t = \pi - \pi/6 = 5\pi/6$ gives $t = 5/6$ s (falling crossing). [2 marks — Q2 reasoning and computation]
So in $[0, 2]$ the trigger fires at $t = 1/6$ s and $t = 5/6$ s.
Part (c) — Evaluating the student's claim. The claim is incorrect. The two firings per period are not evenly spaced. Within one period $[0, 2]$, gap $= 5/6 - 1/6 = $ $4/6 = 2/3$ s. The next firing is at $t = 1/6 + 2 = 13/6$ s in the second period, so gap from $5/6$ is $13/6 - 5/6 = $ $8/6 = 4/3$ s. So firings alternate: $2/3$ s, $4/3$ s, $2/3$ s, $4/3$ s, ... The average is indeed $(2/3 + 4/3)/2 = 1$ s, but no two consecutive firings are actually $1$ s apart. [1 mark]
Why? The two solutions of $\sin\theta = 1/2$ in $[0, 2\pi]$ are at $\pi/6$ and $5\pi/6$ — symmetric about the peak at $\pi/2$, not symmetric about the midline. This is a variant of Trap 01 (only finding one solution per cycle / assuming uniform spacing) — the student counted correctly but assumed even spacing without checking. [1 mark]
Total: 8/8.
Band descriptors for marker.
Band 3: Identifies period and counts solutions in (a); finds only one solution in (b) (Trap 01); accepts the student's "every 1 second" claim. ≈ 3-4 marks.
Band 4: Both solutions in (b); identifies the spacing in (c) is uneven but computes the gaps incorrectly. ≈ 5-6 marks.
Band 5: Full solution including correct gaps and explicit note that the average is $1$ s but no consecutive pair is. ≈ 7 marks.
Band 6: All of the above plus the geometric reasoning (symmetry about the peak, not the midline) and explicit Trap 01 reference. 8/8.