Mathematics Advanced • Year 11 • Module 2 • Lesson 13
Solving Trigonometric Equations Graphically
Build procedural fluency in counting solutions of $\sin x = k$, $\cos x = k$, $\tan x = k$ in a domain — and in extracting the exact solutions where possible.
1. Quick recall
Three warm-ups. 1 mark each
Q1.1 Complete the graphical solution principle:
To solve $f(x) = g(x)$, sketch $y = $ ________ and $y = $ ________ on the same axes; the solutions are the ________-coordinates of the ________ points.
Q1.2 Periodic count: in any interval covering $n$ complete periods of $\sin x$, how many solutions does $\sin x = k$ have if $-1 < k < 1$ and $k \neq 0$?
_____________ solutions per period × $n$ periods = _____________ total.
Q1.3 Boundary value: what happens to the count if $k = 1$ exactly?
2. Worked example — solving $\sin x = 0.5$ in $[0, 2\pi]$
Problem. Solve $\sin x = 0.5$ for $x \in [0, 2\pi]$.
Step 1 — Sketch $y = \sin x$ and $y = 0.5$ on the same axes for $[0, 2\pi]$.
Reason: solutions are $x$-coordinates of intersections.
Step 2 — Find the first-quadrant solution.
$\sin(\pi/6) = 1/2$, so $x_1 = \pi/6$.
Reason: $\pi/6 = 30^\circ$ is a known exact-value angle.
Step 3 — Use symmetry to find the second solution in $[0, 2\pi]$.
Sine is positive in Q1 and Q2; reflection of $\pi/6$ in $x = \pi/2$ gives $\pi - \pi/6 = 5\pi/6$.
Reason: $\sin(\pi - \theta) = \sin\theta$ — the supplementary-angle identity.
Step 4 — Check count against expectation.
2 solutions per period (since $|k| < 1$), 1 period in $[0, 2\pi]$, so 2 total. ✓
Conclusion. $x = $ $\pi/6$ and $5\pi/6$.
3. Faded example — fill in the missing steps
Solve $\cos x = -0.5$ for $x \in [0, 2\pi]$. 4 marks
Step 1 — Sketch $y = \cos x$ and $y = -0.5$ on $[0, 2\pi]$.
Cosine is negative in quadrants ________ and ________.
Step 2 — Reference angle from $\cos \alpha = 1/2$.
$\alpha = $ ________.
Step 3 — Apply quadrant identities.
In Q2: $x = \pi - \alpha = $ ________.
In Q3: $x = \pi + \alpha = $ ________.
Step 4 — Count check.
2 solutions per period (since $|k| < 1$), 1 period in $[0, 2\pi]$, so ________ total. ✓
Conclusion. $x = $ ________ and ________.
4. Graduated practice
For "count" questions just state the number. For "solve" questions list every exact solution.
Foundation — count solutions (4 questions)
| Q | Equation and domain | Number of solutions |
|---|---|---|
| 4.1 1 | $\sin x = 0.3$ on $[0, 2\pi]$ | |
| 4.2 1 | $\cos x = -0.8$ on $[0, 4\pi]$ | |
| 4.3 1 | $\tan x = 1$ on $[0, 2\pi)$ | |
| 4.4 1 | $\sin x = 2$ on $[0, 2\pi]$ |
Standard — solve in $[0, 2\pi]$ (6 questions)
Give exact solutions where possible; otherwise to 3 dp.
4.5 Solve $\sin x = \sqrt{3}/2$ in $[0, 2\pi]$. 2 marks
4.6 Solve $\cos x = -\sqrt{2}/2$ in $[0, 2\pi]$. 2 marks
4.7 Solve $\tan x = -1$ in $[0, 2\pi]$. 2 marks
4.8 Solve $2\sin x = 1$ in $[0, 2\pi]$. (Rewrite first.) 2 marks
4.9 Solve $\cos x = 0$ in $[0, 2\pi]$. 2 marks
4.10 Solve $\sin x = \cos x$ in $[0, 2\pi]$ by dividing by $\cos x$. (Use $\tan x = 1$.) 2 marks
Extension — extended domains and transformed equations (2 questions)
4.11 How many solutions does $\sin x = -0.5$ have in $[-2\pi, 4\pi]$? Justify with a period count. 3 marks
4.12 How many solutions does $\sin(2x) = 0.5$ have in $[0, 2\pi]$? Justify with a period count, then find one exact solution. 3 marks
5. Self-check the easy 3
Tick the first three once you've verified each one.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Graphical solution principle
Sketch $y = f(x)$ and $y = g(x)$ on the same axes; solutions are the $x$-coordinates of the intersection points.
Q1.2 — Count per period
2 solutions per period × $n$ periods = $2n$ total (for $-1 < k < 1$, $k \neq 0$).
Q1.3 — Boundary $k = 1$
The line $y = 1$ touches the sine curve at its maximum (rather than crossing it), giving only 1 solution per period (the tangent point at $x = \pi/2 + 2n\pi$).
Q3 — Faded example: $\cos x = -0.5$
Cosine is negative in Q2 and Q3. Reference angle $\alpha = \pi/3$ (since $\cos(\pi/3) = 1/2$). Q2: $\pi - \pi/3 = 2\pi/3$. Q3: $\pi + \pi/3 = 4\pi/3$. Count: 2 per period × 1 period = 2 total. ✓ $x = 2\pi/3$ and $4\pi/3$.
Q4.1 — $\sin x = 0.3$ on $[0, 2\pi]$
2 solutions (one period, $0 < 0.3 < 1$).
Q4.2 — $\cos x = -0.8$ on $[0, 4\pi]$
2 per period × 2 periods = 4 solutions.
Q4.3 — $\tan x = 1$ on $[0, 2\pi)$
$\tan$ has period $\pi$, so 2 branches; each crosses $y = 1$ once. 2 solutions (at $\pi/4$ and $5\pi/4$).
Q4.4 — $\sin x = 2$
0 solutions. $\sin x$ is bounded by $[-1, 1]$ and never reaches $2$. (Trap-worthy if students try to "solve" anyway.)
Q4.5 — $\sin x = \sqrt{3}/2$
$\sin(\pi/3) = \sqrt{3}/2$, so first solution $\pi/3$. Second (Q2): $\pi - \pi/3 = 2\pi/3$. $x = \pi/3, 2\pi/3$.
Q4.6 — $\cos x = -\sqrt{2}/2$
Reference angle $\pi/4$. Q2: $\pi - \pi/4 = 3\pi/4$. Q3: $\pi + \pi/4 = 5\pi/4$. $x = 3\pi/4, 5\pi/4$.
Q4.7 — $\tan x = -1$
$\tan$ is negative in Q2 and Q4. Reference angle $\pi/4$. Q2: $\pi - \pi/4 = 3\pi/4$. Q4: $2\pi - \pi/4 = 7\pi/4$. $x = 3\pi/4, 7\pi/4$.
Q4.8 — $2\sin x = 1$
Rewrite: $\sin x = 1/2$. $x = \pi/6, 5\pi/6$ (same as the worked example).
Q4.9 — $\cos x = 0$ on $[0, 2\pi]$
$\cos x = 0$ at $x = \pi/2 + n\pi$. In $[0, 2\pi]$: $x = $ $\pi/2$ and $3\pi/2$.
Q4.10 — $\sin x = \cos x$
Divide both sides by $\cos x$ (valid where $\cos x \neq 0$, which it isn't at the answers): $\tan x = 1$. Solutions in $[0, 2\pi]$: $x = \pi/4$ and $5\pi/4$.
Q4.11 — Domain $[-2\pi, 4\pi]$ has length $6\pi$ = 3 periods
$\sin x = -0.5$ has 2 solutions per period; 3 periods, so $2 \times 3 = 6$ solutions. (Explicit list: in $[0, 2\pi]$ they are at $7\pi/6$ and $11\pi/6$; subtract $2\pi$ to get the two in $[-2\pi, 0]$; add $2\pi$ to get the two in $[2\pi, 4\pi]$.)
Q4.12 — $\sin(2x) = 0.5$ on $[0, 2\pi]$
Period of $\sin(2x)$ = $\pi$, so $[0, 2\pi]$ contains 2 full periods. 2 solutions per period × 2 periods = 4 solutions. One exact solution: $2x = \pi/6 \Rightarrow x = \pi/12$. (The full set is $\pi/12,\ 5\pi/12,\ 13\pi/12,\ 17\pi/12$.)