Not all trigonometric equations are easy to solve algebraically — especially when different trig functions are mixed together or when the equation involves transformations. In this lesson, you will learn how to use graphs to find approximate solutions, count the number of solutions in a given interval, and verify algebraic answers by visual inspection.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Consider the equation $\sin x = 0.5$. You know that one solution is $x = 30^\circ$. But because the sine graph repeats forever, there must be infinitely many solutions. How would you find all of them? And if you restricted the domain to $0^\circ \leq x \leq 360^\circ$, how many solutions would there be?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: tan(θ) = sin(θ) + cos(θ).
Right: tan(θ) = sin(θ)/cos(θ); it is a ratio, not a sum.
📚 Core Content
To solve a trigonometric equation graphically, rewrite it so that one side is a trigonometric function and the other side is a constant or another function. Then sketch both graphs on the same axes and find their points of intersection.
Draw $y = \sin x$ and $y = 0.5$ on the same axes. In the interval $0 \leq x \leq 2\pi$, the horizontal line $y = 0.5$ cuts the sine curve twice: once in the first quadrant and once in the second quadrant. The solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.
Draw $y = \cos x$ and $y = -0.5$. In $0 \leq x \leq 2\pi$, the line cuts the cosine curve twice: in the second and third quadrants. The solutions are $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$.
The number of solutions to a trig equation in a given interval equals the number of intersections between the relevant graphs in that interval.
The sine graph completes two full cycles in $4\pi$. The horizontal line $y = 0.3$ cuts each cycle twice. Therefore, there are $2 \times 2 = 4$ solutions.
The tangent graph has period $\pi$, so there are two branches in $[0, 2\pi)$. Each branch intersects $y = 1$ exactly once. Therefore, there are 2 solutions: $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.
🧮 Worked Examples
🧪 Activities
1 $\cos x = 0.5$ for $0 \leq x \leq 2\pi$
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Answer in your workbook.
2 $\sin x = -0.5$ for $0 \leq x \leq 2\pi$
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Answer in your workbook.
3 $\tan x = \sqrt{3}$ for $0 \leq x \leq 2\pi$
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Answer in your workbook.
1 $\sin x = 0.3$ for $0 \leq x \leq 4\pi$
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Answer in your workbook.
2 $\cos x = -0.8$ for $0 \leq x \leq 2\pi$
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Answer in your workbook.
3 $\sin x = 2$ for $0 \leq x \leq 2\pi$
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Answer in your workbook.
Earlier you were asked about $\sin x = 0.5$.
In $0^\circ \leq x \leq 360^\circ$, the horizontal line $y = 0.5$ cuts the sine graph twice: at $x = 30^\circ$ and $x = 150^\circ$. Because sine repeats every $360^\circ$, the general solution is $x = 30^\circ + 360^\circ n$ or $x = 150^\circ + 360^\circ n$ for any integer $n$. So there are infinitely many solutions overall, but only two in one full cycle.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. By sketching the graphs of $y = \sin x$ and $y = \cos x$ on the same axes, find all solutions to $\sin x = \cos x$ in $0 \leq x \leq 2\pi$. 3 MARKS
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Answer in your workbook.
9. How many solutions does $\sin x = 0.7$ have in $0 \leq x \leq 6\pi$? Explain your reasoning. 2 MARKS
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Answer in your workbook.
10. The equation $\cos x = 0.5$ has two solutions in $0 \leq x \leq 2\pi$. By considering the graph of $y = \cos x$, explain what happens to the number of solutions if the domain is extended to $0 \leq x \leq 4\pi$. 3 MARKS
Type your answer below:
Answer in your workbook.
1. $x = \frac{\pi}{3}, \frac{5\pi}{3}$
2. $x = \frac{7\pi}{6}, \frac{11\pi}{6}$
3. $x = \frac{\pi}{3}, \frac{4\pi}{3}$
1. 4 solutions (2 cycles $\times$ 2 intersections)
2. 2 solutions
3. 0 solutions ($y = 2$ is above the maximum of sine)
1. A — $\sin x = 0.5$ has 2 solutions in $[0, 2\pi]$.
2. A — $\sin x = \cos x$ where $\tan x = 1$, at $\frac{\pi}{4}$ and $\frac{5\pi}{4}$.
3. A — $\sin x = 2$ has no solutions.
4. A — 3 cycles $\times$ 2 = 6 solutions.
5. A — $\cos x = -0.5$ at $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$.
Q8 (3 marks): Sketch both graphs [1]. Intersections occur where $\tan x = 1$ [1]. Solutions: $x = \frac{\pi}{4}, \frac{5\pi}{4}$ [1].
Q9 (2 marks): Sine has period $2\pi$, so $6\pi$ contains 3 cycles [1]. Each cycle intersects $y = 0.7$ twice, so there are 6 solutions [1].
Q10 (3 marks): In $[0, 2\pi]$, $y = 0.5$ cuts $y = \cos x$ twice [1]. Cosine has period $2\pi$, so in $[0, 4\pi]$ the pattern repeats [1]. The number of solutions doubles to 4 [1].
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