The algebra on this sub-topic is a factorising exercise students can already do. The marks go on the two things either side of it: turning two ratios into one, and knowing how many answers the interval wants.
Start with the slider at the far right, all four crossings filled, and ask how many solutions the equation has. Someone will say four. Then drag it to 6.283 and ask again. The equation did not change.
The hollow circles are the point: the two excluded crossings stay on screen, drawn empty, so the class can see what has been left out rather than watching it disappear. Ask what a student who stopped at 2π has actually done, and the honest answer is that they answered a different question.
Then drag it slowly from zero and read the count aloud as it goes 0, 1, 2, 3, 4. The jump from 0 to 1 at x = 2.094 is worth pausing on, because it is also where the first solution comes from, and students who think of solutions as a property of the equation rather than of the equation-and-interval find that genuinely surprising.
| Question | Answer |
|---|---|
| 1. The larger solution of 2 sin x = 1 | 5π/6 = 2.618. |
| 2. The rejected root | cos x = 3, impossible. |
| 3. How many on 0 to 4π | 4. |
| 4. The largest | 10π/3 = 10.472. |
| 5. How you know x = π is wrong | B. Substituting gives −4, not 0. |
Question 5 is the one worth discussing as a class. The other wrong options are all plausible-sounding and all wrong for interesting reasons, particularly A: a cosine certainly can be −1, and that is precisely why the error survives.
1 markThe substitution, sin²x = 1 − cos²x.
1 markSolving the quadratic.
1 markRejecting cos x = 3, with the reason.
1 markThe first solution.
1 markThe rest of them in the interval.
Point out the arithmetic of that to a class: the factorising is one mark of five. Students put all their effort into the step they find hardest and lose marks on the two lines they think are not worth writing. The rejection line in particular takes eight words.
| They wrote | What happened |
|---|---|
| 2 on question 3 | The error this page exists for. Worked one cycle out of habit. Drag the slider in front of them rather than explaining it. |
| Only 2.094 and 4.189 | Same error, shown in the solutions rather than the count. Ask them to write "4π is two cycles" at the top of the question next time. |
| 0.524 on question 1 | Gave the calculator's answer. The inverse sine returns one value and the question said solve, not evaluate. |
| 3.665 on question 1 | π + π/6: used the tangent's symmetry on a sine. Its sine is −0.5, so substituting catches it. This mix-up is worth drilling: sine partners with π − x, cosine with 2π − x, tangent adds π. |
| 5.760 on question 1 | 2π − π/6: the cosine's symmetry on a sine. Same diagnosis. |
| −0.5 on question 2 | Rejected the root they should keep. Usually means they have not checked either against the range, just picked the odd-looking one. |
| −1.5 on question 2 | They are working from 2c² + 5c + 3 = 0, so there is a sign slip at the tidy-up step. Worth finding, because its other root is −1 and that one looks legal. |
| x = π | The sign slip, carried through. Nothing about it looks wrong, which is the argument for substituting one answer back every time. |
| 12.566 on question 4 | Gave the end of the interval. The curve is at 6 there. They have confused the largest solution with the largest x. |
"How do I know how many to expect?" Count cycles and multiply. Each legal value of sin x or cos x gives two solutions per 2π, and each legal value of tan x gives one per π, which is also two per 2π. So over whole cycles the count is even for sine and cosine, and a student who writes an odd number on a whole-cycle interval has lost one. The exceptions are the boundary values: sin x = 1 and cos x = ±1 give one per cycle, because the two coincide.
"Can I just graph it?" On Paper 2, yes, and it is good practice to graph first and solve second so you know the count before you start. On Paper 1 there is no graph, which is why the interval-as-cycles habit has to be automatic. Say also that a graphical answer gives decimals and this question's answers are exact multiples of π, which the mark scheme will want.
"Why did we reject 3 but not −0.5?" Because the quadratic knows nothing about cosines. It is a quadratic in a letter, and the letter happens to stand for something confined to −1 to 1. Solving it faithfully produces both roots and the mathematics of the original problem then discards one. That distinction between the substituted equation and the original is exactly what question 5 is testing.
"Does the interval ever exclude the first solution?" Yes, and it is worth one example. On 3 ≤ x ≤ 4π the answer is 4.189, 8.378 and 10.472, three solutions, an odd number, because the interval does not start at a cycle boundary. Give that one after the main example, not before, or the cycle-counting rule will arrive looking unreliable.
| Demonstrate | Graph it over 0 to 12.6 and count the crossings before solving anything. Four. Then do the algebra and get four. The agreement is the point: the graph tells you how many and the algebra tells you which, and neither does the other's job. |
|---|---|
| Where they stick | The default window. Starting at −10 shows crossings at negative x that are not in the interval, and cuts off 10.472 at the other end. Students then report five, or three, and blame the machine. Set the window from the interval, every time. |
| The check | Substitute one solution into the original expression and get zero. On the Nspire that is one line; it catches the sign slip, the wrong symmetry and a mistyped root all at once. |
The root finder needs bounds around a single crossing. With four on screen students put the bounds around two and get whichever the machine prefers, which looks arbitrary and is not.
| Step | What |
|---|---|
| 1 | 2 sin x = 1 on 0 to 2π. Collect the single answer, then ask for all of them. |
| 2 | The three symmetries, as a table: sine π − x, cosine 2π − x, tangent add π. |
| 3 | The same equation on 0 to 4π, so the add-a-period step appears on something easy. |
| 4 | Now the guide's question. Two ratios, so nothing can be solved. Let them notice. |
| 5 | The substitution, and the quadratic. Insist on the positive leading coefficient. |
| 6 | Both roots, then the rejection, in writing. |
| 7 | Four solutions. Then the figure, and the slider down to 2π. |
| 8 | The sign-slip version, as a spot-the-error: cos x = −1 and x = π. Substitute and get −4. |
Do not say "there are always two solutions per cycle". It is nearly always true and the exceptions are examinable: cos x = 1 gives one per cycle, and a tangent equation gives one per π. A rule with unmentioned exceptions is worse than a method, and the method here is to find one solution and use the symmetry, which handles every case including the awkward ones.
Do not let the rejection pass as obvious. Every year a class nods at "cos x = 3 is impossible" and then half of them omit the line in the exam, because it felt too obvious to write. It is a mark. Make them write it in class, every time, until it is reflex, and tell them plainly that the examiner cannot award what is not on the paper.