When an equation like $\sin x = \tfrac{x}{3}$ resists algebraic attack, a graph rescues you. Sketch $y = \sin x$ and $y = \tfrac{x}{3}$ on the same axes, every crossing is an exact solution. In this lesson you'll master reading intersection multiplicity, estimating values from graphs, and knowing when to trust the geometry over the algebra.
Consider the equation $\cos x = x$. Without algebra sketch the two graphs ($y = \cos x$ and $y = x$) in your head or on paper and estimate how many times they cross. Write your reasoning below.
To solve $f(x) = g(x)$ graphically, rearrange (if needed) so both sides are functions you can sketch, then count the intersections of $y = f(x)$ and $y = g(x)$. Each crossing point gives one solution.
Step 1, Rearrange. Write the equation in the form $f(x) = g(x)$ where each side is a recognisable function.
Step 2, Sketch. Draw both curves accurately on the same axes, noting key features (amplitude, period, asymptotes, intercepts).
Step 3, Count. Each intersection = one solution. Read off approximate $x$-values if required.
To determine the number of solutions to a trigonometric equation, rewrite it as $f(x) = g(x)$ and sketch both curves on the same set of axes over the given domain.
Key principle: the number of solutions equals the number of intersection points.
Why does the number depend on the coefficient?
Method: rewrite as f(x) = g(x), sketch both curves, count intersection points; Each crossing = 1 solution; tangency = 1 repeated solution (not 2)
Pause, copy the graphical counting method into your book: rewrite the equation as $f(x) = g(x)$, sketch both curves on the same axes, count crossings (each = 1 solution), note that tangency gives one repeated solution, not two.
Quick check: How many solutions does $\sin x = 2$ have?
We just saw the graphical method: rewrite as $f(x) = g(x)$, sketch both, count crossings. That raises a question: when the trig function has an amplitude $a$ or period multiplier $b$, how do these transformations change the number of intersections, and what conditions make solutions impossible? This card answers it → amplitude $a$ means no solutions when $|c| > |a|$; larger $b$ compresses the period and doubles the crossings per domain.
Understanding how transformations shift and stretch trig curves lets you predict intersection counts before picking up a pencil.
Amplitude changes ($y = a\sin x$): Increasing $a$ stretches the wave vertically. A larger amplitude means the wave can reach further, potentially creating more intersections with a horizontal line $y = c$.
Period changes ($y = \sin(bx)$): A larger $b$ compresses the wave horizontally, so within a fixed domain it completes more cycles and can intersect a line many more times.
Amplitude a: vertical stretch. Solutions possible only when |c| |a|.; Period multiplier b: horizontal compression. Doubles the cycles doubles the solutions per domain.
Pause, copy the transformation-effect rules into your book: amplitude $a$ → no solutions when $|c| > |a|$; period multiplier $b$ → each doubling of $b$ doubles the solution count in a fixed domain.
Did you get this? True or false: The equation $\sin(2x) = 0.5$ has twice as many solutions on $[0, 2\pi]$ as $\sin x = 0.5$.
Worked examples · 3 in a row, reveal as you go
How many solutions does $2\sin x = x$ have for all real $x$? Justify with a sketch.
Find the number of solutions to $\cos x = \dfrac{x}{2}$ for $-\pi \leq x \leq \pi$.
How many solutions does $\sin(3x) = 0.4$ have for $0 \leq x \leq 2\pi$?
Fill the gap: The equation $\sin(2x) = 0.6$ on $[0, 2\pi]$ has solutions, because $\sin(2x)$ completes 2 full cycles on this interval.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: If a line is tangent to $y = \sin x$ at exactly one point in $[0, 2\pi]$, then the equation has exactly one solution in that interval.
Activities · practice with the ideas
Sketch $y = \sin x$ and $y = 0.8$ on the same axes for $0 \leq x \leq 2\pi$. How many intersection points are there?
How many solutions does $\cos x = x^2 - 1$ have for all real $x$? Justify by sketching both curves.
Determine the number of solutions to $\sin(2x) = \dfrac{x}{\pi}$ for $-\pi \leq x \leq \pi$.
For what range of values of $k$ does $\sin x = k$ have exactly 2 solutions in $[0, 2\pi]$?
Explain why $\tan x = \sin x$ has infinitely many solutions, but all solutions satisfy either $\cos x = 1$ or $\sin x = 0$. (Hint: manipulate the equation algebraically.)
Odd one out: Three of these statements about graphical solutions are correct. Which one is WRONG?
Earlier you estimated how many times $y = \cos x$ and $y = x$ intersect.
The exact answer is 1 solution near $x \approx 0.739$ (the Dottie number). At $x = 0$, $\cos 0 = 1 > 0$, so cosine starts above; for large positive $x$, $\cos x \leq 1 < x$, so the line eventually wins. For large negative $x$, $x < -1 \leq \cos x$. The two curves cross exactly once, near $x \approx 0.739$. Did your intuition land close?
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. How many solutions does $\sin x = 0.5$ have for $0 \leq x \leq 4\pi$? (1 mark)
Q2. By sketching $y = \cos x$ and $y = 1 - x$ on the same axes for $0 \leq x \leq \pi$, determine the number of solutions to $\cos x = 1 - x$. (2 marks)
Q3. The equation $\sin x = kx$ (where $k > 0$) has exactly one solution when $k \geq 1$. Explain why, and find the approximate value of $k$ at which $\sin x = kx$ transitions from 3 solutions to 1 solution. (3 marks)
Activity answers:
1. $y=\sin x$ meets $y=0.8$ at 2 points per period, so on $[0,4\pi]$ (2 full periods): 4 intersections.
2. $y=\cos x$ (between $-1$ and $1$) and $y=x^2-1$ (parabola with vertex at $(0,-1)$): sketch shows 3 intersections near $x \approx 0, \pm 1.17$.
3. $\sin(2x)$ has period $\pi$, so on $[-\pi,\pi]$ it completes 2 full cycles. Line $y=x/\pi$ (slope $1/\pi \approx 0.318$) is relatively shallow. Graphical analysis gives 5 intersections.
4. $\sin x = k$ has exactly 2 solutions on $[0,2\pi]$ when $-1 < k < 1$ and $k \neq 1, -1$ (boundaries give 1 solution). So $k \in (-1, 0) \cup (0, 1)$. (At $k=0$, two solutions: $x=0,\pi,2\pi$, actually 3, so $k \neq 0$ gives 2 solutions for $0 < |k| < 1$.)
5. $\tan x = \sin x \Rightarrow \sin x/\cos x = \sin x \Rightarrow \sin x(1 - \cos x)/\cos x = 0 \Rightarrow \sin x = 0$ or $\cos x = 1$. Both give $x = n\pi$.
Q1 (1 mark): On $[0, 4\pi]$, $\sin x$ completes 2 full periods. Each period contributes 2 crossings of $y=0.5$. Total: 4 solutions [1].
Q2 (2 marks): At $x=0$: $\cos 0 = 1 = 1-0 = 1$, the curves meet [1]. At $x=\pi$: $\cos\pi = -1$ and $1-\pi \approx -2.14$, cosine is above line. On $(0,\pi)$, $\cos x$ starts at 1, drops to $-1$, while line drops from 1 to $-2.14$. The curves start together at $x=0$ and by the end cosine is above the line, they meet at $x=0$ and cross once more near $x \approx 1.26$. So 2 solutions on $[0,\pi]$ [1].
Q3 (3 marks): For $k \geq 1$, the slope of $kx$ at the origin is $k \geq 1$, which equals or exceeds the maximum slope of $\sin x$ (which is 1 at $x=0$). So the line is at least as steep as $\sin x$'s steepest point, meaning the line cannot be "outrun" by the sine curve, it crosses only at $x = 0$ [1]. The transition occurs when $y = kx$ is tangent to $y = \sin x$ at some $x_0 > 0$: need $k = \sin x_0 / x_0$ and $k = \cos x_0$ simultaneously, giving $\tan x_0 = x_0$. The smallest positive solution is $x_0 \approx 4.49$, so $k \approx \cos(4.49) \approx -0.217$, but since $k > 0$, the transition is at $k \approx 1$ (the sine's derivative condition at origin). So the answer is $k \geq 1$ gives 1 solution [1, 1].
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