- A trigonometric equation such as \(\sin x = 0.6\) usually has many solutions, because sine, cosine and tangent repeat. An exam always gives an interval, and you must find every solution in it.
- Method: find the principal value with your calculator, use symmetry (the CAST diagram or the graph) to find a second value, then add or subtract whole periods.
- For \(\sin(2x – 30°) = k\), solve for \(u = 2x – 30°\) over the adjusted interval first, then convert back to \(x\).
- In our case, a daylight model \(D(t) = 12 + 2\sin\!\left(\frac{2\pi(t – 80)}{365}\right)\) gives more than 13 hours of daylight from about day 111 to day 232: 122 days.
- Exam point: set your calculator to degrees or radians to match the interval, and give answers in the same unit.
The case: when are the days longer than 13 hours?
A school in the northern hemisphere runs an after-school outdoor sports club. The coach wants to schedule it only on days with more than 13 hours of daylight, so that sessions finish before dark. She models the number of daylight hours on day \(t\) of the year (with \(t = 1\) on 1 January) by
The model says daylight averages 12 hours, swings 2 hours above and below that, repeats every 365 days, and crosses 12 hours rising on day 80, which is near the March equinox. (The model is illustrative. Real day lengths depend on latitude and are better taken from astronomical tables, but a sine curve is a standard first model for them.)
The question is: for which days is \(D(t) > 13\)? To answer it we must solve \(D(t) = 13\), a trigonometric equation, and find every solution in the year, not just the first one the calculator shows. That is the skill this post teaches.
The concept: why trig equations have many solutions
Solving \(2x + 1 = 7\) gives one answer. Solving \(\sin x = 0.5\) does not. The sine graph rises and falls forever, and it crosses the line \(y = 0.5\) twice in every cycle. So in \(0° \le x \le 360°\) there are two answers, 30° and 150°, and in \(0° \le x \le 720°\) there are four.
Your calculator’s \(\sin^{-1}\) button gives only one value, called the principal value. For \(\sin^{-1}\) and \(\tan^{-1}\) it lies between −90° and 90°; for \(\cos^{-1}\) it lies between 0° and 180°. Everything else comes from two facts about the graphs:
- Symmetry. \(\sin(180° – x) = \sin x\), \(\cos(360° – x) = \cos x\) and \(\tan(180° + x) = \tan x\). So each principal value has a partner within the same cycle.
- Periodicity. Sine and cosine repeat every 360° (or \(2\pi\)); tangent repeats every 180° (or \(\pi\)).[1] Adding or subtracting a whole period gives more solutions.
Many students use the CAST diagram to remember the signs. Divide a circle into four quadrants. In the first quadrant (0° to 90°) All three ratios are positive; in the second, only Sine; in the third, only Tangent; in the fourth, only Cosine. If \(\sin x\) is positive, the solutions lie in the first and second quadrants.[4]
Here \(k\) is the given value, \(\alpha\) is the principal value from the calculator (\(\alpha = \sin^{-1}k\), \(\cos^{-1}k\) or \(\tan^{-1}k\)), and \(n\) is any whole number: 0, ±1, ±2 and so on. In radians, replace 180° with \(\pi\) and 360° with \(2\pi\). For sine and cosine there are no solutions at all if \(|k| > 1\), because the graphs never go above 1 or below −1.
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You do not have to memorise the general formulas. Sketching the graph over the interval and reading off where it meets the line \(y = k\) is just as reliable, and it shows you how many solutions to expect before you start.[3]
Why this method fits the case
Daylight rises and falls over the year and then repeats, which is exactly the behaviour of a sine curve.[4] The model has the general form \(y = A + B\sin(C(t – h))\), where \(A\) is the average, \(B\) is the amplitude, \(\frac{2\pi}{C}\) is the period and \(h\) is the shift. Once the model is set up, the question “when is daylight more than 13 hours?” becomes “where is the sine curve above the line \(y = 13\)?” That needs both crossing points, which is precisely what the method gives.
The method also tells you how many solutions to look for. The argument \(\frac{2\pi(t – 80)}{365}\) runs over just under one full cycle as \(t\) goes from 0 to 365, so we expect at most two crossings. Without this check it is easy to stop after the first answer.
When is it not the right tool? A pure sine model is only an approximation. Real daylight curves are not perfectly symmetric, and near the poles the model breaks down completely, with weeks of 24-hour daylight or darkness. For accurate planning you would use published sunrise and sunset tables. Also, some equations, such as \(\sin x = x/2\), mix a trig term with a non-trig term; they cannot be solved with inverse trig functions and need a numerical method or a graph instead.
Setting it up
Here are the quantities in the case.
| Quantity | Symbol | Value |
|---|---|---|
| Day of the year | \(t\) | \(0 \le t \le 365\) |
| Average daylight | \(A\) | 12 hours |
| Amplitude | \(B\) | 2 hours |
| Period | – | 365 days |
| Day daylight crosses 12 h, rising | \(h\) | 80 |
| Target | – | \(D(t) = 13\) hours |
| Angle in radians | \(\theta\) | \(\dfrac{2\pi(t – 80)}{365}\) |
When \(t = 0\), \(\theta = \frac{2\pi(-80)}{365} \approx -1.377\). When \(t = 365\), \(\theta = \frac{2\pi(285)}{365} \approx 4.906\). So we need solutions for \(\theta\) in the interval \(-1.377 \le \theta \le 4.906\). The model uses radians, so the calculator must be in radian mode.
Solving it
We start with a basic equation, then a compound angle, then the daylight case.
Solve \(\sin x = 0.6\) for \(0° \le x \le 360°\), giving answers to 1 decimal place.
\(x = 36.9°\) or \(x = 143.1°\).
Solve \(\sin(2x – 30°) = 0.5\) for \(0° \le x \le 360°\).
Four solutions: \(x = 30°, 90°, 210°, 270°\). The “2x” doubles the number of cycles, so it doubles the number of answers.
Find the days on which the model gives exactly 13 hours of daylight, and hence the period with more than 13 hours.
The model gives more than 13 hours of daylight on days 111 to 232 inclusive: 122 days, roughly late April to late August.
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What the answer tells us
The club can run on 122 days of the year. The window is centred on day 171.25, where \(\theta = \frac{\pi}{2}\) and daylight peaks at 14 hours. That is near the June solstice, as we would expect. The two crossing days, 110.4 and 232.1, are the same distance (60.8 days) either side of the peak, because the sine curve is symmetric about its maximum.
Reasonableness check. A sine curve spends one third of each cycle above half its amplitude: \(\sin\theta > 0.5\) for \(\theta\) between \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\), which is \(\frac{2\pi}{3}\), one third of \(2\pi\). One third of 365 days is about 121.7 days. Our answer of 122 whole days matches.
Sensitivity. If the coach relaxed the target to 12.5 hours, the equation becomes \(\sin\theta = 0.25\), with principal value 0.2527. The crossings move to about day 94.7 and day 247.8, giving a window of about 153 days. Lowering the target by just half an hour adds about a month. Because the curve is flattest near its peak and steepest near 12 hours, small changes to a target near the average shift the dates more slowly than changes to a target near the top. Raising the target to 14.5 hours would give \(\sin\theta = 1.25\), which has no solution: the model never reaches 14.5 hours.
The chart shows the model across the year, with the 13-hour target as a dashed line. The club’s season is the stretch where the blue curve lies above the line.
For a full treatment of trigonometric graphs, exact values and equations in degrees and radians, see the Mathematics Made Simple textbook. To practise finding every solution in an interval until it is routine, the Mathematics Made Simple Practice Questions book has graded sets with full worked answers.
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Practice questions
Sketch a quick graph for each one so you know how many answers to expect, then solve.
EasyQ1. Solve \(\cos x = 0.3\) for \(0° \le x \le 360°\), to 1 decimal place. Show worked answer
Cosine is positive in quadrants 1 and 4, so the other solution is \(360° – 72.54° = 287.46°\).
\(x = 72.5°\) or \(x = 287.5°\).
EasyQ2. Solve \(3\tan x = 2\) for \(0° \le x \le 360°\), to 1 decimal place. Show worked answer
Tangent repeats every 180°: \(33.69° + 180° = 213.69°\).
\(x = 33.7°\) or \(x = 213.7°\).
MediumQ3. Solve \(\tan 2x = \sqrt{3}\) for \(0° \le x \le 360°\). Show worked answer
\(\tan u = \sqrt{3}\) gives \(u = 60°\), then add 180° each time: \(60°, 240°, 420°, 600°\).
Halve: \(x = 30°, 120°, 210°, 300°\).
MediumQ4. Solve \(2\cos^2 x + \sin x – 1 = 0\) for \(0° \le x \le 360°\). Show worked answer
Factorise: \((2\sin x + 1)(\sin x – 1) = 0\).
\(\sin x = 1\) gives \(x = 90°\). \(\sin x = -\frac{1}{2}\) gives \(x = 210°\) or \(x = 330°\).
So \(x = 90°, 210°, 330°\).
ExamQ5. (a) Show that the equation \(3\sin^2 x + 5\cos x = 5\) can be written as \(3\cos^2 x – 5\cos x + 2 = 0\). [2] (b) Hence solve \(3\sin^2 x + 5\cos x = 5\) for \(0 \le x \le 2\pi\), giving answers in radians to 3 significant figures where not exact. [4] Show worked answer
(b) Factorise: \((3\cos x – 2)(\cos x – 1) = 0\) [M1].
\(\cos x = 1\): \(x = 0\) or \(x = 2\pi\) [A1].
\(\cos x = \frac{2}{3}\): \(x = \cos^{-1}\frac{2}{3} = 0.841\) [A1] and \(x = 2\pi – 0.841 = 5.44\) [A1].
Solutions: \(x = 0,\ 0.841,\ 5.44,\ 2\pi\).
Frequently asked questions
How do I know how many solutions to expect?
Count the cycles in the interval. Sine and cosine usually give two solutions per 360° cycle, and tangent gives one per 180°. If the angle is \(2x\), there are twice as many cycles in the same range of \(x\), so twice as many solutions. A rough sketch of the graph confirms the count.
What is the CAST diagram?
It is a memory aid showing which ratios are positive in each quadrant: All in the first, Sine in the second, Tangent in the third and Cosine in the fourth. It helps you place the second solution, but the graph of the function tells you the same thing.
Should I work in degrees or radians?
Use whatever the interval uses. If the interval is \(0 \le x \le 2\pi\) with no degree sign, answer in radians and set your calculator to radian mode. Mixing units is one of the most common ways to lose marks.
Why does an equation like sin x = 1.4 have no solutions?
Because \(\sin x\) is never greater than 1 or less than −1. The line \(y = 1.4\) lies above the whole sine curve, so they never meet. Say this in words in an exam rather than leaving the root out silently.
References
- Abramson, J. et al. (2021) Algebra and Trigonometry 2e. Houston: OpenStax. https://openstax.org/details/books/algebra-and-trigonometry-2e
- Cambridge International (2024) Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. https://www.cambridgeinternational.org/programmes-and-qualifications/cambridge-international-as-and-a-level-mathematics-9709/
- International Baccalaureate Organization (2019) Mathematics: analysis and approaches guide. Geneva: IBO.
- Abramson, J. et al. (2021) Precalculus 2e. Houston: OpenStax. https://openstax.org/details/books/precalculus-2e
