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General solutions of circular equations and interval extraction

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General solutions of circular equations and interval extraction

A general solution describes every angle that satisfies a circular equation. The form depends on the function's period and its symmetric positions

\[\begin{gathered}-1\le a\le 1,\quad \sin\alpha=a\\\sin x=a\Rightarrow x=2n\pi+\alpha\text{ or }x=(2n+1)\pi-\alpha,\quad n\in\mathbb{Z}\end{gathered}\]
  1. 1Identify the period Sine and cosine repeat after a full turn, while tangent repeats after a half turn \(\displaystyle P_{\sin}=2\pi,\quad P_{\cos}=2\pi,\quad P_{\tan}=\pi\)
  2. 2Find the exact base angle Use the exact circular value to identify the coefficient of π \(\displaystyle \sin\left(\frac{\pi}{6}\right)=\frac{1}{2}\)
  3. 3Include symmetric positions Sine and cosine need both positions when the target is not an endpoint value \(\displaystyle \cos x=\frac{1}{2}\Rightarrow x=2n\pi\pm\frac{\pi}{3}\)
  4. 4State the integer parameter The integer parameter generates all repeated turns of the solution \(\displaystyle \tan x=1\Rightarrow x=n\pi+\frac{\pi}{4},\quad n\in\mathbb{Z}\)
Year 12, VCE Units 3 & 4, Chapter 1, 4 difficulty levels, No calculatorBuild a worksheet

General solutions of circular equations and interval extraction

A general solution describes every angle that satisfies a circular equation. The form depends on the function's period and its symmetric positions

\[\begin{gathered}-1\le a\le 1,\quad \sin\alpha=a\\\sin x=a\Rightarrow x=2n\pi+\alpha\text{ or }x=(2n+1)\pi-\alpha,\quad n\in\mathbb{Z}\end{gathered}\]
  1. Identify the period Sine and cosine repeat after a full turn, while tangent repeats after a half turn \(\displaystyle P_{\sin}=2\pi,\quad P_{\cos}=2\pi,\quad P_{\tan}=\pi\)
  2. Find the exact base angle Use the exact circular value to identify the coefficient of π \(\displaystyle \sin\left(\frac{\pi}{6}\right)=\frac{1}{2}\)
  3. Include symmetric positions Sine and cosine need both positions when the target is not an endpoint value \(\displaystyle \cos x=\frac{1}{2}\Rightarrow x=2n\pi\pm\frac{\pi}{3}\)
  4. State the integer parameter The integer parameter generates all repeated turns of the solution \(\displaystyle \tan x=1\Rightarrow x=n\pi+\frac{\pi}{4},\quad n\in\mathbb{Z}\)

Worked example

Solve the circular equation for x, giving the general solution

\[\cos\left(2x+\frac{\pi}{6}\right)=-\frac{1}{2}\]
  1. \(\displaystyle u=2x+\frac{\pi}{6},\quad \cos u=-\frac{1}{2}\)Use u for the inner argument before mapping the solutions back to x
  2. \(\displaystyle P=2\pi,\quad \alpha=\frac{2\pi}{3},\quad u=2n\pi\pm\frac{2\pi}{3}\)Cosine has period 2π and two symmetric positions for this target
  3. \(\displaystyle 2x+\frac{\pi}{6}=2n\pi+\frac{2\pi}{3}\Rightarrow x=n\pi+\frac{\pi}{4}\)Subtract the angle added to 2x, then divide both sides by 2 for the first branch
  4. \(\displaystyle 2x+\frac{\pi}{6}=2n\pi-\frac{2\pi}{3}\Rightarrow x=n\pi-\frac{5\pi}{12}\)Apply the same mapping to the second branch so both families are included

Common mistakes

Using the wrong period
✕\(\displaystyle \tan x=1\Rightarrow x=2n\pi+\frac{\pi}{4}\)✓\(\displaystyle \tan x=1\Rightarrow x=n\pi+\frac{\pi}{4}\)
Tangent repeats after π, not 2π, so a full-turn period misses valid solutions
Omitting a symmetric branch
✕\(\displaystyle \sin x=\frac{1}{2}\Rightarrow x=2n\pi+\frac{\pi}{6}\)✓\(\displaystyle \sin x=\frac{1}{2}\Rightarrow x=2n\pi+\frac{\pi}{6}\text{ or }x=(2n+1)\pi-\frac{\pi}{6}\)
A sine target between -1 and 1 usually occurs at two symmetric positions in each full turn
Using the positive base angle for a negative target
✕\(\displaystyle \sin x=-\frac{1}{2}\Rightarrow \alpha=\frac{\pi}{6}\)✓\(\displaystyle \sin x=-\frac{1}{2}\Rightarrow \alpha=-\frac{\pi}{6}\)
With the negative-angle convention, a negative sine target uses a negative base-angle coefficient

Difficulty levels

introWrite the general family for a basic equation with a positive exact target
standardWrite both families for a negative target or handle an affine argument and a positive interval
challengeMap an affine argument through a mixed interval or test a stated member
examSelect the only correct family and then give one exact interval-solution coefficient

Questions about this topic

What are the periods of sine, cosine and tangent?+

Sine and cosine have period 2π. Tangent has period π, so its solution family uses nπ rather than 2nπ

Why do sine and cosine usually have two solution branches?+

Their graphs have two symmetric positions with the same value in one full turn, except at endpoint values such as 1 or -1

What does n ∈ Z mean in a general solution?+

It means n can be any integer. Changing n adds or subtracts complete periods and generates every solution

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 1: Solving equations

In this chapter

Factorise, solve and reconstruct a rational quadraticSolve a quadratic exactly with the quadratic formulaSolve an equation that reduces to quadratic formDetermine parameter values from a discriminant conditionDivide polynomials and state quotient and remainderApply the remainder and factor theoremsMove between roots, factors, polynomial rules and equationsDetermine parameters in a polynomial identityUse unit-circle coordinates to evaluate circular functionsEvaluate exact circular-function values by symmetry and periodicityUse exact trigonometric identities and quadrant informationSolve circular equations over finite domainsGeneral solutions of circular equations and interval extractionLocate and bisect a circular-equation rootRearrange literal equations and recognise equivalent line formsSolve a pair of simultaneous linear equationsSolve a linear system with three or four unknownsVerify coordinate incidence and determine a parameterSolve literal equations and simultaneous literal equationsClassify and parameterise linear systems