Topic 3.8 · AA Standard Level

Two answers or four, and the question decides

2sin²x + 5cos x + 1 = 0 has two solutions on 0 to 2π and four on 0 to 4π. Same equation, same algebra, different answer. The interval is not decoration and it is where the marks are.

0 to 12.6
4solutions in the interval
2.09, 4.19, 8.38, 10.47which ones
all fouryou have found

The curve never changes. Only the band does, and the number of answers is a property of the band and the curve together.

Start with the easy one, so the pattern is visible

Solve 2 sin x = 1 for 0 ≤ x ≤ 2π.

  1. sin x = ½.
  2. The calculator gives x = 0.524, which is π/6. That is one answer and the question asked for all of them in the interval.
  3. sin(π − x) = sin x, so π − π/6 = 5π/6 = 2.618 has the same sine.
  4. Both are inside 0 to 2π, so the answer is x = π/6 or 5π/6.

That is the whole method in miniature: get one solution, use the symmetry to get its partner, then add whole periods until you leave the interval. The calculator does step 2 and nothing else.

Which symmetry depends on the function. Sine: the partner is π − x. Cosine: the partner is −x, or 2π − x inside the interval. Tangent: there is no partner, because its period is π and not 2π, so you just add π. Getting this wrong is the single commonest way to lose half the solutions.

The guide's own question, which is a quadratic in disguise

Solve 2sin²x + 5cos x + 1 = 0 for 0 ≤ x ≤ 4π.

Two ratios appear, so nothing can be solved yet. The Pythagorean identity from 3.6 turns one into the other:

  1. Substitute sin²x = 1 − cos²x: 2(1 − cos²x) + 5cos x + 1 = 0.
  2. Tidy up: −2cos²x + 5cos x + 3 = 0, so 2cos²x − 5cos x − 3 = 0. Multiplying by −1 to get a positive leading coefficient is worth the line.
  3. It factorises: (2cos x + 1)(cos x − 3) = 0, giving cos x = −½ or cos x = 3.
  4. Reject cos x = 3. A cosine is never above 1, so that root of the quadratic is not a solution of the equation. Say so in writing; it is a mark.
  5. cos x = −½ gives x = 2π/3 = 2.094 and x = 4π/3 = 4.189 in the first cycle.
  6. The interval goes to 4π, which is two cycles. Add 2π to each: 8.378 and 10.472. Both are under 4π = 12.566, so both count.

Four solutions: 2.094, 4.189, 8.378 and 10.472. Drag the slider down to 6.283 and watch two of them leave the band. Nothing about the algebra changed.

The two places this question is lost, and neither is the algebra.

There is a third, quieter one. A sign slip at step 2 gives 2cos²x + 5cos x + 3 = 0, whose roots are −1 and −1.5. The −1.5 is rejected as usual, and −1 is a perfectly legal cosine, so it produces x = π and looks like a tidy answer. Substituting back gives −4, not 0. Check one solution in the original equation and this whole family of errors is caught.

Graphically, and when to do it that way

The same question answered by graph: plot y = 2sin²x + 5cos x + 1 over the interval and count where it crosses zero. The figure is that graph, and the crossings are the four answers.

Use the graph when the equation will not factorise, and use the algebra when it will. A quadratic in one ratio always factorises or yields to the formula, so this one is algebra; something like sin x = x/3 is not, and then the graph is the only route and the answer is a decimal.

Either way the interval sets the window. Plot 0 to 12.566 and the four crossings are all on screen. Plot the default −10 to 10 and you will see some of them, miss one, and have no way of knowing.

On the GDC: counting solutions in an interval

This is one of the few places where the calculator is genuinely better than the algebra, not for finding the answers but for knowing how many there are. Graph it over the interval and the count is visible before you solve anything.

When you may use it. Analysis Paper 1 is non-calculator, and the quadratic-in-disguise questions live there because the answers are exact multiples of π. On Paper 2 graph it first to count, then solve algebraically to earn the method marks.

TI-Nspire CX II

  1. doc → Settings → Document Settings → Angle → Radian. In degrees the interval 0 to 12.566 is 12.6 degrees out of a 360-degree cycle, so the window holds about a twenty-ninth of one wave and the curve looks almost flat
  2. A Graphs page with f1(x)=2*(sin(x)) x² +5*cos(x)+1. Use the x² key: typing ^2 opens a superscript box and the rest of the expression goes inside it
  3. Set the window from the interval. menu → Window / Zoom → Window Settings, x from 0 to 12.6 and y from −5 to 7. Now count the crossings: four
  4. menu → Analyze Graph → Zero, with bounds either side of each crossing: 2.094, 4.189, 8.378, 10.472. Four separate searches, one per crossing

Casio fx-CG50

  1. SHIFT MENU SET UP → Angle → Rad
  2. MENU → Graph with Y1=2(sin(X)) x² +5cos(X)+1, then F6 DRAW
  3. SHIFT F3 V-Window and type Xmin 0, Xmax 12.6, Ymin −5, Ymax 7. F3 STD starts at −10 and hides the shape of the interval
  4. SHIFT F5 G-SOLVE → ROOT gives the leftmost crossing in the window, 2.094; the right arrow steps right through the other three

The mark people lose. Giving the solutions from one cycle when the interval covers two. It is worth saying how this looks to an examiner: two correct values and two missing is not a half-understood answer, it is an answer to a different question. The habit: before solving, write the interval as a number of cycles. 0 to 4π is two. Then you know to expect double whatever the first cycle gives.

Your turn

1. Solve 2 sin x = 1 for 0 ≤ x ≤ 2π. Give the larger solution, to 3 decimal places.

2. For 2sin²x + 5cos x + 1 = 0, the substitution gives a quadratic in cos x. Give the root that has to be rejected.

3. How many solutions does 2sin²x + 5cos x + 1 = 0 have on 0 ≤ x ≤ 4π?

4. Give the largest of those solutions, to 3 decimal places.

5. A student solves the quadratic and gets cos x = −1, then reports x = π. How do you know that is wrong without redoing the algebra?

Question 5. A student gets cosine x equals minus 1 and reports x equals pi. How do you know that is wrong without redoing the algebra?
Where the marks go

1 markThe substitution, turning two ratios into one.

1 markSolving the quadratic.

1 markRejecting the impossible root, with the reason.

1 markThe first solution.

1 markAll the others in the interval.

Five marks and only one of them is the quadratic. The substitution and the rejection are each worth as much as the algebra, and the last mark is worth as much again, so a student who solves it beautifully and stops at 2π gets four of five at best.

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