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.
$\cos x = \tfrac{1}{2}$ 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.
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:
$n$ ranges over all integers $\mathbb{Z}$. The restriction tells you which values of $n$ to keep.
Given a basic equation $f(\theta) = k$ where $|k|\le 1$ (for $\sin$/$\cos$), the general solution captures every angle on the real line.
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.
=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 $\cos\theta = \tfrac{\sqrt{3}}{2}$?
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:
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!
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
Find all solutions of $\sin x = \dfrac{1}{\sqrt{2}}$ for $-2\pi \le x \le 2\pi$.
Find all solutions of $\sin(2x - \tfrac{\pi}{6}) = \tfrac{1}{2}$ for $0 \le x \le \pi$.
How many solutions does $\tan(3x) = \sqrt{3}$ have in $0 \le x \le 2\pi$?
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
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 $0\le u\le 4\pi$.
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?
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).
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Write the general solution of $\sin x = \dfrac{\sqrt{3}}{2}$ and find all solutions in $[0, 2\pi]$. (2 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)
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)
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].
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