Year 11 SAC HSC: Solve $\sin(2x) = 1/2$ for $x \in [0, 2\pi]$.

Q: Year 11 SAC HSC: Solve $\sin(2x) = 1/2$ for $x \in [0, 2\pi]$.

Let $u = 2x$. Then $u \in [0, 4\pi]$ and $\sin u = 1/2$. Solutions for $\sin u = 1/2$ in $[0, 4\pi]$: $u = \pi/6, 5\pi/6, 13\pi/6, 17\pi/6$. Back-substitute $u = 2x$: $x = u/2 = \pi/12, 5\pi/12, 13\pi/12, 17\pi/12$. Markers reward the substitution, the expanded interval for $u$, the four solutions in that interval, and back-substitution.

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VICMath MethodsSyllabus dot point

How are circular functions extended to model periodic phenomena?

Sketch and analyse trigonometric functions $y = a\sin(b(x - h)) + k$ and $y = a\cos(b(x - h)) + k$, identifying amplitude, period, phase and vertical translation, and solve trig equations over a specified interval

A focused answer to the VCE Maths Methods Unit 2 dot point on circular functions. Sketches transformed sine and cosine graphs, identifies amplitude $|a|$, period $2\pi/|b|$, phase shift $h$ and vertical translation $k$, and works the VCAA SAC-style trig equation $\sin(2x) = 1/2$ on a stated interval.

Generated by Claude OpusReviewed by Better Tuition Academy5 min answerUpdated 2026-05-19

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What this dot point is asking

VCAA wants you to sketch and analyse transformed sine and cosine functions, identifying amplitude, period, phase shift and vertical translation, and to solve trigonometric equations over a specified interval.

Parent functions

$\sin x$ and $\cos x$ :

Domain: all real $x$ . Range: $[-1, 1]$ .
Period: $2\pi$ . Amplitude: $1$ .

Transformed form IMATH_6

Parameter	Effect
IMATH_7	Amplitude. Range $[k - \\|a\\|, k + \\|a\\|]$ . If $a < 0$ , vertical reflection.
IMATH_10	Period = $2\pi/\\|b\\|$ . Larger $\\|b\\|$ compresses horizontally.
IMATH_13	Phase shift; graph moves $h$ right (if positive).
IMATH_15	Vertical translation; centre line $y = k$ .

Same parameters apply to $y = a\cos(b(x - h)) + k$ .

Solving trig equations over a stated interval

Isolate the trig function.
Find the principal solution.
Use symmetry and periodicity to find all solutions in the stated interval.
If the equation involves $bx$ , list all solutions for $bx$ in the expanded interval and divide by $b$ .

Worked example

Solve $2\cos(x) + 1 = 0$ for $x \in [0, 2\pi]$ .

$\cos x = -1/2$ . Reference angle $\pi/3$ . Cosine is negative in Q2 and Q3.

$x = \pi - \pi/3 = 2\pi/3$ , or $x = \pi + \pi/3 = 4\pi/3$ .

Common traps

Confusing $b$ with the period. Period is $2\pi/b$ , not $b$ .

Forgetting to expand the interval. When solving $\sin(2x) = c$ for $x \in [0, 2\pi]$ , you must find all solutions for $2x$ in $[0, 4\pi]$ before dividing by $2$ .

Missing solutions outside the principal range. $\sin$ and $\cos$ are periodic; there are infinitely many solutions, but only those in the stated interval count.

In one sentence

Transformed circular functions $y = a\sin(b(x - h)) + k$ have amplitude $|a|$ , period $2\pi/|b|$ , phase shift $h$ and centre line $y = k$ ; solving trig equations over an interval uses the reference angle plus quadrant symmetry, and equations involving $bx$ require finding solutions for $bx$ in the expanded interval before dividing by $b$ .

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

Year 11 SAC4 marksSolve $\sin(2x) = 1/2$ for $x \in [0, 2\pi]$.

Show worked answer →

Let $u = 2x$ . Then $u \in [0, 4\pi]$ and $\sin u = 1/2$ .

Solutions for $\sin u = 1/2$ in $[0, 4\pi]$ : $u = \pi/6, 5\pi/6, 13\pi/6, 17\pi/6$ .

Back-substitute $u = 2x$ : $x = u/2 = \pi/12, 5\pi/12, 13\pi/12, 17\pi/12$ .

Markers reward the substitution, the expanded interval for $u$ , the four solutions in that interval, and back-substitution.