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Solve circular equations over finite domains

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What should you do first with the equation?

\[2\sin(x)+1=0\]
Year 11, VCE Units 1 & 2, Chapter 10, 4 difficulty levels, Calculator allowedBuild a worksheet

Solve a circular equation on a finite domain

First isolate the circular-function value, then generate all symmetric and periodic solutions. Keep only the values in the stated interval and count a shared endpoint once

\[T_{\sin}=T_{\cos}=2\pi,\quad T_{\tan}=\pi\]
  1. Isolate the function Rearrange the equation until the sine, cosine or tangent value is alone \(\displaystyle 2\sin x+1=0\Rightarrow\sin x=-\frac{1}{2}\)
  2. Check the function range A sine or cosine value must lie between −1 and 1 before solutions are kept \(\displaystyle -1\le\sin x\le1\)
  3. Include symmetric solutions Use the relevant quadrant pattern and period to list every angle with the target value \(\displaystyle \sin x=\frac{1}{2},\quad x=\frac{\pi}{6},\frac{5\pi}{6}\text{ on }[0,2\pi)\)
  4. Filter the stated interval Retain only solutions in the given domain and do not list a shared endpoint twice \(\displaystyle x\in[0,2\pi)\Rightarrow\text{retain each solution once}\)

Worked example

Solve the equation on the stated domain

\[3\cos x-\frac{3}{2}=0,\quad0\le x<2\pi\]
  1. \(\displaystyle 3\cos x=\frac{3}{2}\)Add the constant term to both sides
  2. \(\displaystyle \cos x=\frac{1}{2}\)Divide both sides by 3
  3. \(\displaystyle \alpha=\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)This gives the reference angle
  4. \(\displaystyle x=\frac{\pi}{3},\quad2\pi-\frac{\pi}{3}=\frac{5\pi}{3}\)Cosine is positive in quadrants I and IV, and both angles lie in the domain

Common mistakes

Applying the inverse too early
✕\(\displaystyle 2\sin x+1=0\Rightarrow x=\sin^{-1}(0)\)✓\(\displaystyle 2\sin x+1=0\Rightarrow\sin x=-\frac{1}{2}\)
The circular-function value must be isolated before applying an inverse function
Omitting a symmetric solution
✕\(\displaystyle \sin x=\frac{1}{2},\quad x=\frac{\pi}{6}\)✓\(\displaystyle \sin x=\frac{1}{2},\quad x=\frac{\pi}{6},\frac{5\pi}{6}\text{ on }[0,2\pi)\)
Sine has the same positive value in quadrants I and II, so both angles are needed in one period
Counting a boundary twice
✕\(\displaystyle \sin x=0\text{ on }[0,2\pi]\Rightarrow x=0,0,\pi,2\pi\)✓\(\displaystyle \sin x=0\text{ on }[0,2\pi]\Rightarrow x=0,\pi,2\pi\)
The shared endpoint represents one angle and must be counted once

Difficulty levels

introSolve an isolated sine, cosine or tangent equation on a standard finite interval
standardRearrange an equation and apply boundary or approximate-angle checks in the domain
challengeSolve an equation involving a square, quadratic, ratio or affine argument
examSolve the full equation, including all valid branches, periods and domain restrictions

Questions about this topic

Why can a circular equation have more than one solution?+

Sine, cosine and tangent repeat and can take the same value at different angles, so every matching angle in the stated domain must be listed

How does the interval affect the answer?+

The interval decides which periodic solutions are retained. An endpoint included by the interval is counted once, even if it is also reached by the next cycle

Why must the circular function be isolated first?+

Inverse functions and quadrant patterns apply to the circular-function value. Rearranging first prevents the coefficient or constant from being treated as part of the angle

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 10: Transformed circular functions

In this chapter

Solve circular equations over finite domainsTransform and reconstruct sine and cosine functionsFeatures and transformations of tangent functionsForm and use periodic sine and cosine modelsUse exact trigonometric identities and quadrant information