Blog - Page 13 - AMC 12 Mathematics Competition Website

Dividing both sides of an equation can make it shorter, but it can also remove valid solutions. In AMC 12 trigonometry practice, check whether the expression you want to divide by could be zero before cancelling it.

The Mathematical Association of America's AMC overview includes trigonometry within AMC 12's high school mathematics scope and excludes calculus. This guide uses an original equation and unit-circle diagram to practise a useful habit: keep every valid branch, then check the required interval. It is a teaching example, rather than an official AMC 12 question or a difficulty benchmark.

Know the identities but still miss solutions? Try the example below before reading the explanation. Compare your solution list with the four points on the unit circle.

To discuss what to work on next, scan the contact QR code or contact Hanlin on WeChat. Send AMC12 TRIG with your school grade and preparation question.

An original equation to solve

How many distinct real values of x satisfy

2 sin(x) cos(x) = cos(x), where 0 ≤ x < 2π?

(A) 2 (B) 3 (C) 4 (D) 5 (E) 6

All angles in this article are in radians. The interval includes 0 and excludes 2π. List the actual angles before counting them; that makes omissions and repeated values easier to spot.

Move everything to one side and factor

Subtract cos(x) from both sides:

2 sin(x) cos(x) - cos(x) = 0.

Factor out the shared term without dividing by it:

cos(x) [2 sin(x) - 1] = 0.

A product is zero when at least one factor is zero. We therefore need to solve both branches: cos(x) = 0 or sin(x) = 1/2.

Solve both branches on the unit circle

On the unit circle, the point at angle x has coordinates (cos(x), sin(x)). Cosine is the horizontal coordinate and sine is the vertical coordinate.

cos(x) = 0: the horizontal coordinate is zero at the top and bottom of the circle. In the required interval, x = π/2 or 3π/2.
sin(x) = 1/2: the horizontal line at height 1/2 meets the circle in the first and second quadrants. The angles are x = π/6 or 5π/6.

Combine the two lists and put them in order:

π/6, π/2, 5π/6, 3π/2.

All four lie in the interval, and none is repeated. There are 4 distinct solutions, so the answer is (C).

A unit circle showing four solutions of two sine x cosine x equals cosine x: pi over six, pi over two, five pi over six and three pi over two. — Blue points satisfy cos(x) = 0; green points satisfy sin(x) = 1/2. The two branches together give all four solutions. Original practice diagram.

Why dividing by cos(x) loses two solutions

If you immediately divide the original equation by cos(x), you obtain 2 sin(x) = 1. That step assumes cos(x) is not zero, so it only describes part of the original problem.

At x = π/2 and x = 3π/2, cosine is zero. Substituting either angle into the original equation gives 0 = 0. Both angles are valid, but division would have discarded them.

You can divide after separating and checking the zero case. Factoring is convenient here because it displays both cases in one line. The goal is to preserve the original solution set throughout the calculation.

Check that the list is complete

The factorisation is an identity, so any solution must belong to at least one of the two branches. Within one full turn, cos(x) = 0 gives exactly two angles, and sin(x) = 1/2 gives exactly two. Conversely, every listed angle makes at least one factor zero and therefore satisfies the original equation.

This checks both directions: no valid angle was missed, and no extra angle was introduced. Substitution is useful for testing candidates, but checking a few candidates alone does not establish that the list is complete.

Use a short checklist for trigonometric equations

Record the interval and angle unit. Distinguish 0 ≤ x < 2π from a closed interval, and keep radians separate from degrees.
Check restrictions before transforming. Before dividing by a function, consider its zeros. For equations involving fractions, record where denominators are undefined.
Solve every factor or branch. Use the unit circle to include all relevant quadrants.
Check the range. For real angles, sine and cosine lie between -1 and 1. An equation such as sin(x) = 3/2 has no real solution.
Filter, combine and verify. Apply the interval, remove duplicate values and test the original equation.

Try two short practice questions

Solve cos(x) [2 sin(x) - 3] = 0 for 0 ≤ x < 2π. Which branch can be ruled out immediately?
Solve sin(x) [2 cos(x) - 1] = 0 for 0 ≤ x < 2π. How would the number of distinct solutions change if the interval included 2π?

Show the answers and reasoning

Question 1: The branches are cos(x) = 0 and sin(x) = 3/2. The second branch is impossible for real x. The solutions are π/2 and 3π/2, giving 2 solutions.

Question 2: The branches are sin(x) = 0 and cos(x) = 1/2. On 0 ≤ x < 2π, their combined solution list is 0, π/3, π, 5π/3, giving 4 solutions.

On the closed interval 0 ≤ x ≤ 2π, the extra value 2π also satisfies the equation, so there are 5 solutions. Although 0 and 2π point in the same direction on the circle, they are distinct real values of x. The question asks for angle values, so the stated interval matters.

Turn a missed solution into a useful review note

Record the exact step where the solution set changed. "Check cos(x) = 0 before dividing" gives you an action to repeat. "Be more careful with trigonometry" does not identify the cause.

For your next session, solve a similar equation with a different sine or cosine value, then explain why your list is complete. Return to the AMC 12 preparation guide if you need help organising broader topic review. The MAA's K-12 educator resources provide official preparation materials and sample competition resources for further practice.

Discuss your AMC 12 preparation with Hanlin

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Sources: the official MAA AMC overview and K-12 educator resources linked above. The examples, diagram and practice suggestions were created for this article.

In AMC 12 preparation, a quadratic equation becomes more demanding when a coefficient is an unknown parameter. Instead of solving for one numerical answer, you must identify all parameter values that make a condition true. The discriminant provides a useful route, but the boundary cases still need attention.

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An original parameter question

For which real values of k does x² − 2kx + k + 2 = 0 have at least one real solution for x? This is an original teaching problem, not an official AMC 12 question.

The coefficient of x² is always 1, so the equation remains quadratic for every real k. This check is important in parameter problems: if the leading coefficient could become zero, that special case would need separate treatment.

Translate the condition into an inequality

For a quadratic ax² + bx + c = 0 with real coefficients and a ≠ 0, real roots exist when b² − 4ac ≥ 0. Here a = 1, b = −2k and c = k + 2.

The discriminant is (−2k)² − 4(k + 2) = 4k² − 4k − 8 = 4(k − 2)(k + 1). Therefore we need (k − 2)(k + 1) ≥ 0.

Solve the inequality by signs

The critical values are −1 and 2. Below −1, both factors are negative, so their product is positive. Between −1 and 2, the factors have opposite signs, so the product is negative. Above 2, both factors are positive.

At either critical value the product is zero, which is allowed by the phrase “at least one real solution”. The complete answer is k ≤ −1 or k ≥ 2.

Do not write −1 ≤ k ≤ 2 simply because the factors vanish at the endpoints. The sign between the endpoints must be checked. Substituting k = 0 gives x² + 2 = 0, which has no real roots and immediately exposes that incorrect interval.

Check what happens at the boundaries

At k = −1, the equation is x² + 2x + 1 = (x + 1)² = 0. There is one repeated real root, x = −1. At k = 2, it becomes x² − 4x + 4 = (x − 2)² = 0, with repeated root x = 2.

If the question instead required two distinct real roots, the discriminant would need to be strictly positive. The answer would then be k < −1 or k > 2. A small change in wording changes whether the boundary values are included.

Confirm the result by completing the square

Rewrite the original equation as (x − k)² = k² − k − 2 = (k − 2)(k + 1). A square of a real number cannot be negative, so the right side must be nonnegative. This gives the same condition and explains its meaning without relying only on a memorised discriminant rule.

For your next review, record the leading-coefficient check, the required inequality, the sign intervals and the boundary interpretation. Use MAA's official resources for additional AMC practice. Understanding why endpoints are included is more reliable than memorising a final interval.

New to the subject? Read our AMC 12 Preparation: What to Study and How to Practise Effectively.

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