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Module 7 · L17 of 20 ~40 min ⚡ +100 XP available

General Solutions with Restrictions

The general solution to $\sin\theta = k$ is $\theta = n\pi + (-1)^n\alpha$ — it produces infinitely many angles. In HSC questions you're handed a restriction like $0 \le x \le 4\pi$ and must sift that infinite list down to a finite one. This lesson teaches the systematic approach: write the full general form, substitute integers, stop when you leave the interval.

Today's hook — $\cos x = \tfrac{1}{2}$ has infinitely many solutions on $\mathbb{R}$. Before doing the lesson, guess: how many solutions do you expect in the interval $0 \le x \le 4\pi$? Jot your answer now and check it after card 06.

0/5QUESTS

Recall — gut answer first

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$\cos x = \tfrac{1}{2}$ on $\mathbb{R}$ has infinitely many solutions. Without solving — how many solutions do you expect in the restricted interval $0 \le x \le 4\pi$? Write your estimate and reasoning below.

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The core strategy

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Every "restricted interval" question has two stages: first, write the general solution (infinitely many), then substitute integer values of $n$ and keep only those inside the given interval.

The three standard general solutions are:

$\sin\theta = k \;\Rightarrow\; \theta = n\pi + (-1)^n\alpha$ where $\sin\alpha = k$, $\alpha\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$
$\cos\theta = k \;\Rightarrow\; \theta = 2n\pi \pm \alpha$ where $\cos\alpha = k$, $\alpha\in[0,\pi]$
$\tan\theta = k \;\Rightarrow\; \theta = n\pi + \alpha$ where $\tan\alpha = k$, $\alpha\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})$

$n$ ranges over all integers $\mathbb{Z}$. The restriction tells you which values of $n$ to keep.

$\theta = 2n\pi \pm \alpha, \quad n\in\mathbb{Z}$

Start with $n=0$

Then try $n=1,-1,2,-2,\ldots$ in turn. Stop as soon as both branches leave the interval.

Two branches for $\cos$ and $\sin$

Remember the $\pm$ gives two expressions for each $n$. Check both independently — either can be in or out of the interval.

Count, then list

Many HSC questions ask "how many solutions?" — you only need to count valid values of $n$, not necessarily evaluate each angle.

What you'll master

Know

Key facts

The three general solution formulae for $\sin$, $\cos$, $\tan$
Principal values: $\alpha\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$ for $\sin$; $\alpha\in[0,\pi]$ for $\cos$; $\alpha\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})$ for $\tan$
What it means to "restrict to an interval"

Understand

Concepts

Why substituting consecutive integers into the general form generates all solutions
How the period of each function determines spacing of solutions
The relationship between the interval width and the number of solutions

Can do

Skills

Write the general solution for any basic trig equation
Find all solutions in a given interval by systematic substitution
Count solutions without listing every one (for wide intervals)

Key terms

General solutionAn expression (involving integer $n$) that gives every solution of a trig equation on $\mathbb{R}$.

Principal value ($\alpha$)The unique angle in the principal range that satisfies $\sin\alpha=k$ (or $\cos\alpha=k$, $\tan\alpha=k$).

Restricted intervalA specific domain such as $[0,2\pi]$ or $[-\pi,\pi]$ within which all valid solutions must lie.

PeriodThe length of one full cycle: $2\pi$ for $\sin$ and $\cos$; $\pi$ for $\tan$. Solutions repeat every period.

$n\in\mathbb{Z}$$n$ is any integer $(\ldots,-2,-1,0,1,2,\ldots)$. Each integer gives one or two candidate solutions.

Valid solutionA candidate from the general form that satisfies the inequality defining the restricted interval.

The three general solution forms

core concept

Given a basic equation $f(\theta) = k$ where $|k|\le 1$ (for $\sin$/$\cos$), the general solution captures every angle on the real line.

$$\sin\theta = k \;\Longrightarrow\; \theta = n\pi + (-1)^n\alpha, \quad n\in\mathbb{Z}$$

$$\cos\theta = k \;\Longrightarrow\; \theta = 2n\pi \pm \alpha, \quad n\in\mathbb{Z}$$

$$\tan\theta = k \;\Longrightarrow\; \theta = n\pi + \alpha, \quad n\in\mathbb{Z}$$

Here $\alpha$ is always the principal value — the angle you read off your calculator (or exact table) in the appropriate range.

Why does the $(-1)^n$ appear in the sine formula? Because $\sin$ is positive in Q1 and Q2 but the pattern of solutions alternates: $\alpha, \pi-\alpha, 2\pi+\alpha, 3\pi-\alpha, \ldots$ The $(-1)^n$ sign flip encodes that alternation compactly.

Exam tip. In the HSC you will also encounter equations like $\sin(2x+\tfrac{\pi}{3}) = \tfrac{\sqrt{3}}{2}$. Apply the general form to the compound argument $u = 2x+\tfrac{\pi}{3}$, expand for integer $n$, then isolate $x$ and check the interval.

=k: general solution = n+(-1)^n; principal value =^{-1}k[-{2},{2}]; =k: general solution = 2n; principal value =^{-1}k[0,]

Pause — copy all three general solution forms into your book: $\sin\theta=k \Rightarrow \theta=n\pi+(-1)^n\alpha$; $\cos\theta=k \Rightarrow \theta=2n\pi\pm\alpha$; $\tan\theta=k \Rightarrow \theta=n\pi+\alpha$.

Quick check: What is the general solution of $\cos\theta = \tfrac{\sqrt{3}}{2}$?

Applying a restriction: step-by-step method

core concept

We just saw that general solutions like $\theta = 2n\pi \pm \alpha$ give infinitely many answers. That raises a question: when a question asks for solutions in $[0, 4\pi]$, how do you systematically extract exactly the solutions in that interval from the general formula? This card answers it → substitute $n = 0, 1, 2, \ldots$ (and negative values if needed) into the general formula and list all values that fall within the given bounds.

Once you have the general solution, restrict it to an interval $[a, b]$ as follows:

Write the general solution expression(s).
Set the expression $\ge a$ and $\le b$; solve for the range of integers $n$.
Substitute each valid integer to find the exact solutions.
List all solutions and verify each is within $[a, b]$.

Example: Solve $\cos x = \tfrac{1}{2}$ for $0 \le x \le 4\pi$.

Principal value: $\alpha = \cos^{-1}\tfrac{1}{2} = \tfrac{\pi}{3}$.

General solution: $x = 2n\pi \pm \tfrac{\pi}{3}$, $n\in\mathbb{Z}$.

Branch 1: $x = 2n\pi + \tfrac{\pi}{3}$. For $n=0$: $x=\tfrac{\pi}{3}$ ✓. For $n=1$: $x=\tfrac{7\pi}{3}$ ✓. For $n=2$: $x=\tfrac{13\pi}{3} > 4\pi$ ✗.

Branch 2: $x = 2n\pi - \tfrac{\pi}{3}$. For $n=0$: $x=-\tfrac{\pi}{3} < 0$ ✗. For $n=1$: $x=\tfrac{5\pi}{3}$ ✓. For $n=2$: $x=\tfrac{11\pi}{3}$ ✓. For $n=3$: $x=\tfrac{17\pi}{3} > 4\pi$ ✗.

Solutions: $x = \dfrac{\pi}{3},\, \dfrac{5\pi}{3},\, \dfrac{7\pi}{3},\, \dfrac{11\pi}{3}$ — that's 4 solutions. Check your hook estimate!

Interval width heuristic. Each full period of $2\pi$ contributes 2 solutions for $\cos$ and $\sin$. The interval $[0,4\pi]$ spans exactly 2 periods, so expect $2\times 2 = 4$ solutions — confirmed above. This is a quick sanity check.

x = 1{2}, 0 x 4: ={3}; general x=2n{3}; Enumerate branches: n=0,1,2, until outside interval

Pause — copy the restriction method into your book: from the general solution, substitute $n = 0, 1, 2, \ldots$ (and negative $n$ if needed) until values exceed the interval; list all solutions within the given bounds.

Did you get this? True or false: $\tan x = 1$ has exactly 4 solutions in the interval $0 \le x \le 4\pi$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · SINE RESTRICTION

Find all solutions of $\sin x = \dfrac{1}{\sqrt{2}}$ for $-2\pi \le x \le 2\pi$.

Principal value: $\alpha = \sin^{-1}\!\dfrac{1}{\sqrt{2}} = \dfrac{\pi}{4}$

$\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$ — this is a standard exact value. Principal range for $\sin^{-1}$ is $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$.

PROBLEM 2 · COMPOUND ANGLE

Find all solutions of $\sin(2x - \tfrac{\pi}{6}) = \tfrac{1}{2}$ for $0 \le x \le \pi$.

Let $u = 2x - \tfrac{\pi}{6}$. When $0\le x\le\pi$, we have $-\tfrac{\pi}{6} \le u \le \tfrac{11\pi}{6}$.

Substitute endpoints of $x$ into $u = 2x-\tfrac{\pi}{6}$: at $x=0$, $u=-\tfrac{\pi}{6}$; at $x=\pi$, $u=2\pi-\tfrac{\pi}{6}=\tfrac{11\pi}{6}$.

PROBLEM 3 · COUNTING SOLUTIONS

How many solutions does $\tan(3x) = \sqrt{3}$ have in $0 \le x \le 2\pi$?

Let $u=3x$. Range of $u$: $0\le u\le 6\pi$. $\alpha=\tan^{-1}\!\sqrt{3}=\tfrac{\pi}{3}$. General: $u=n\pi+\tfrac{\pi}{3}$.

The domain of $u$ spans $6\pi$ and $\tan$ has period $\pi$, so there are 6 full periods.

Fill the gap: For $\tan(2x) = 1$ on $0\le x\le 2\pi$, let $u=2x$. The range of $u$ is $0\le u\le$ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the negative-$n$ branch

Students often only try $n=0,1,2,\ldots$ and miss solutions from $n=-1,-2,\ldots$ Whenever the left endpoint $a$ of your interval is negative (or less than $\alpha$), you must try negative integers too.

Trap 02

Not converting the domain for compound arguments

For $\sin(2x-\tfrac{\pi}{3})$, the restriction on $x$ must be converted to a restriction on $u=2x-\tfrac{\pi}{3}$ before applying the general form. Solving for $x$ directly without this step leads to missing solutions or including invalid ones.

Trap 03

Inclusive vs exclusive endpoints

Check whether endpoints are included ($\le$) or excluded ($<$). A solution that equals an endpoint is valid for $\le$ but not for $<$. This distinction can change the solution count by 1 or 2.

Did you get this? True or false: when solving $\cos(2x) = -1$ on $0\le x\le 2\pi$ you should first find the range of $u=2x$ to be $0\le u\le 4\pi$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Write the general solution of $\sin x = -\tfrac{1}{2}$ and find all solutions in $[0, 2\pi]$.

Find all solutions of $\cos(x + \tfrac{\pi}{4}) = 0$ for $0 \le x \le 2\pi$.

How many solutions does $\sin(3x) = \tfrac{\sqrt{3}}{2}$ have in $[0, \pi]$?

Find all solutions of $\tan x = -1$ for $-\pi \le x \le \pi$.

Solve $\cos(2x - \tfrac{\pi}{3}) = -\tfrac{1}{2}$ for $0 \le x \le \pi$. List all exact solutions.

Odd one out: Three of these are solutions of $\cos x = \tfrac{1}{2}$ on $[0, 4\pi]$. Which one is NOT?

Revisit your thinking

Earlier you estimated how many solutions $\cos x = \tfrac{1}{2}$ has in $[0, 4\pi]$.

The answer is exactly 4 solutions: $\tfrac{\pi}{3}, \tfrac{5\pi}{3}, \tfrac{7\pi}{3}, \tfrac{11\pi}{3}$. The key insight is that a $4\pi$-wide interval spans two full periods of $\cos$, and each period contributes 2 solutions (one from each branch of the $\pm$ in the general form).

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Write the general solution of $\sin x = \dfrac{\sqrt{3}}{2}$ and find all solutions in $[0, 2\pi]$. (2 marks)

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ApplyBand 43 marks

Q2. Find all solutions of $\cos\!\left(2x + \dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2}$ for $0 \le x \le \pi$. (3 marks)

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AnalyseBand 53 marks

Q3. How many solutions does $\sin(4x) = \dfrac{1}{2}$ have in the interval $0 \le x \le 2\pi$? Justify your answer. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers: 1. $\sin x=-\tfrac{1}{2}$: $\alpha=-\tfrac{\pi}{6}$; solutions in $[0,2\pi]$: $\tfrac{7\pi}{6}, \tfrac{11\pi}{6}$ · 2. $\cos(x+\tfrac{\pi}{4})=0$: $u\in[\tfrac{\pi}{4},\tfrac{9\pi}{4}]$; $u=\tfrac{\pi}{2},\tfrac{3\pi}{2},\tfrac{5\pi}{2}$; $x=\tfrac{\pi}{4},\tfrac{5\pi}{4},\tfrac{9\pi}{4}-\tfrac{\pi}{4}=\tfrac{9\pi}{4}$ — check: only $x=\tfrac{\pi}{4},\tfrac{5\pi}{4}$ in $[0,2\pi]$ · 3. $u=3x\in[0,3\pi]$: $\sin u=\tfrac{\sqrt{3}}{2}$, $u=\tfrac{\pi}{3},\tfrac{2\pi}{3},\tfrac{\pi}{3}+2\pi=\tfrac{7\pi}{3},\tfrac{2\pi}{3}+2\pi=\tfrac{8\pi}{3}$ — 4 solutions · 4. $\tan x=-1$, $\alpha=-\tfrac{\pi}{4}$; $x=-\tfrac{\pi}{4},\tfrac{3\pi}{4}$ in $[-\pi,\pi]$ · 5. $u=2x-\tfrac{\pi}{3}\in[-\tfrac{\pi}{3},\tfrac{5\pi}{3}]$; $\cos u=-\tfrac{1}{2}$, $\alpha=\tfrac{2\pi}{3}$; $u=\tfrac{2\pi}{3},\tfrac{4\pi}{3}$; $x=\tfrac{\pi}{2},\tfrac{5\pi}{6}$

Q1 (2 marks): $\alpha=\tfrac{\pi}{3}$. General solution: $x=n\pi+(-1)^n\tfrac{\pi}{3}$ [1]. In $[0,2\pi]$: $n=0\Rightarrow\tfrac{\pi}{3}$; $n=1\Rightarrow\tfrac{2\pi}{3}$ [1]. Solutions: $x=\dfrac{\pi}{3}, \dfrac{2\pi}{3}$.

Q2 (3 marks): $u=2x+\tfrac{\pi}{6}$; range $[\tfrac{\pi}{6},\tfrac{13\pi}{6}]$ [1]. $\cos u=-\tfrac{\sqrt{3}}{2}$, $\alpha=\tfrac{5\pi}{6}$; $u=\tfrac{5\pi}{6},\tfrac{7\pi}{6},\tfrac{5\pi}{6}+2\pi=\tfrac{17\pi}{6}>\tfrac{13\pi}{6}$ — 2 valid values [1]. $x=\tfrac{u-\tfrac{\pi}{6}}{2}$: $x=\tfrac{1}{2}(\tfrac{5\pi}{6}-\tfrac{\pi}{6})=\tfrac{\pi}{3}$; $x=\tfrac{1}{2}(\tfrac{7\pi}{6}-\tfrac{\pi}{6})=\tfrac{\pi}{2}$ [1].

Q3 (3 marks): $u=4x\in[0,8\pi]$ [1]. Period $2\pi$: 4 full periods ⇒ 2 solutions per period = 8 solutions total. Or list: $u=\tfrac{\pi}{6},\tfrac{5\pi}{6}$ and each shifted by $2\pi$ for 4 periods [1+1].

Boss battle · The Interval Gatekeeper

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Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering general-solution questions. Lighter alternative to the boss.

Mark lesson as complete

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Module overview · Extension 1 · Checkpoint 4