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.