OMPT drill — Trigonometry

Trigonometric equations

Solving sin(x) = c means finding every solution in the interval, not just the calculator one. The forgotten second solution is the classic lost mark. Keep a unit circle sketch next to you for the first pass, then redo the set without it.

Tested inOMPT-B OMPT-D OMPT-G

Lesson

Solving $sin x = \frac{1}{2}$ is a different kind of task from solving $3 x = 12$ , because sine repeats forever: there are infinitely many angles with sine one-half, and the question decides which of them you owe. Usually it hands you an interval, all solutions in $[0, 2 π)$ , say, and the examiner’s real test is whether you find all of them or stop at the first.

The structure never changes. Within one turn of the circle, $sin x = c$ has two solutions (a height line crosses the circle twice), and so does $cos x = c$ (a vertical line does too). Your calculator, or the exact-value table, hands you exactly one of them. The other comes from symmetry: for sine, the partner of $x$ is $π - x$ ; for cosine, it is $- x$ , or equivalently $2 π - x$ . Forgetting the partner solution is, by a distance, the most common lost mark in this topic, which is why every worked example below hunts for it explicitly.

Problem 1Numeric answer

How many solutions does $2 sin x = 3$ have on the interval $[0, 2 π)$ ?

Show the worked solution

Answer: 2

Rewriting gives $sin x = \frac{3}{2}$ . Sine hits that height twice per cycle: once at $x = \frac{π}{3}$ and once at the supplementary angle $x = π - \frac{π}{3} = \frac{2 π}{3}$ . So there are 2 solutions.

Problem 2Multiple choice

What is the smallest positive solution of $cos 2 x = \frac{1}{2}$ ?

A $\frac{π}{6}$
B $\frac{π}{3}$
C $\frac{2 π}{3}$
D $\frac{5 π}{6}$

Show the worked solution

Answer: A — $\frac{π}{6}$

I solve for the inner angle first: $2 x = \pm \frac{π}{3} + 2 k π$ . The smallest positive choice is $2 x = \frac{π}{3}$ , so $x = \frac{π}{6}$ . Halving at the very end is the whole trick with compressed arguments.

Problem 3Numeric answer

How many solutions does $sin^{2} x - sin x = 0$ have on $[0, 2 π)$ ?

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Answer: 3

Factoring gives $sin x (sin x - 1) = 0$ , so either $sin x = 0$ or $sin x = 1$ . On $[0, 2 π)$ , $sin x = 0$ at $x = 0$ and $x = π$ , and $sin x = 1$ at $x = \frac{π}{2}$ . That is 3 solutions in total.

Problem 4Multiple choice

Which expression gives all solutions of $tan x = 1$ , with $k$ an integer?

A $x = \frac{π}{4} + 2 k π$
B $x = \frac{π}{4} + k π$
C $x = \pm \frac{π}{4} + 2 k π$
D $x = \frac{π}{4} + \frac{k π}{2}$

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Answer: B — $x = \frac{π}{4} + k π$

Tangent has period $π$ , not $2 π$ , so from the base solution $x = \frac{π}{4}$ I step in multiples of $π$ : $x = \frac{π}{4} + k π$ . The option with $2 k π$ misses half the solutions, such as $\frac{5 π}{4}$ .

Problem 5Spot the error

A student solves $sin 2 x = sin x$ on $[0, 2 π)$ . Step 1: rewrite as $2 sin x cos x = sin x$ . Step 2: divide both sides by $sin x$ to get $2 cos x = 1$ . Step 3: solve $cos x = \frac{1}{2}$ , giving $x = \frac{π}{3}$ and $x = \frac{5 π}{3}$ . Which step contains the error?

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Answer: Step 2

Step 1 is a correct double-angle expansion, and Step 3 solves its equation correctly. The damage happens in Step 2: dividing by $sin x$ is only legal when $sin x \neq = 0$ , and it discards the cases $x = 0$ and $x = π$ . Bringing everything to one side and factoring, $sin x (2 cos x - 1) = 0$ , gives the full solution set ${0, \frac{π}{3}, π, \frac{5 π}{3}}$ .

Problem 6Numeric answer

Solve $2 cos^{2} x + 3 cos x - 2 = 0$ on $[0, 2 π)$ , then give the sum of all solutions, rounded to two decimals.

Show the worked solution

Answer: 6.28

Treating $cos x$ as the unknown, the quadratic factors as $(2 cos x - 1) (cos x + 2) = 0$ . The root $cos x = - 2$ is impossible, so only $cos x = \frac{1}{2}$ survives, giving $x = \frac{π}{3}$ and $x = \frac{5 π}{3}$ . Their sum is $2 π \approx 6.28$ .