Mathematical Methods · Units 3 & 4

Solving Circular Function Equations

Learn to solve sine, cosine and tangent equations over a given domain in radians the easy way, with plain English intuition, exact values, symmetry, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

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Spin a point around a circle forever and its height traces a wave that repeats again and again. That endless repeating is exactly why a circular function equation almost never has just one answer. Ask “where is the height equal to a half” and the wave hits that level over and over. Solving these equations is really about hunting down every spot in your domain where the wave lands on the value you want, no more and no fewer.

Why there is never just one answer

A normal equation like 2x=62x = 6 has a single tidy solution. Circular functions are different because they repeat. Sine and cosine repeat every 2π2\pi, and tangent repeats every π\pi. So once you find one angle that works, sliding around the circle by a full period lands you on another angle that works just as well.

That is good news and a trap at the same time. The good news is the pattern is predictable. The trap is that students often stop too early and hand in only the first answer they find. The examiner reports are blunt about this. The single most common mistake is giving only one or two solutions and forgetting the rest that the period demands.

123456 -1-0.50.51 (5π/6, -√3/2) (7π/6, -√3/2)
Solving cos(x) = -√3/2 means finding where the cosine wave crosses the dashed level y = -√3/2; over [0, 2π] it does so twice, at x = 5π/6 and x = 7π/6.

The recipe that finds every solution

You do not guess. There is a reliable four step recipe, and exact values do the heavy lifting.

  1. Get the function on its own, so you have cos⁡(x)=a number\cos(x) = \text{a number}, or sine, or tangent.
  2. Find the reference angle by asking which standard angle gives that value, ignoring the sign for a moment. The values for π6\frac{\pi}{6}, π4\frac{\pi}{4} and π3\frac{\pi}{3} are expected knowledge, so learn them cold.
  3. Use the sign to choose the quadrants. Positive cosine lives in quadrants one and four, negative cosine in two and three. Positive sine in one and two, negative sine in three and four. Tangent is positive in one and three, negative in two and four.
  4. Build every angle in those quadrants that falls inside your domain, then stop.

A clean way to remember the signs is All Students Take Care, reading the quadrants from one to four for where All, Sine, Tangent and Cosine are positive.

The double angle trap

Here is the error that quietly costs the most marks. When the equation reads sin⁡(2x)=…\sin(2x) = \dots the angle inside is 2x2x, not xx. If you solve as if it were plain xx you will only ever find half the solutions.

The fix is a substitution. Let u=2xu = 2x and stretch the domain to match. If xx runs over [0,2π][0, 2\pi], then u=2xu = 2x runs over [0,4π][0, 4\pi], which is two full turns. Solve for uu across that bigger domain, find all of its solutions, then divide each one by two to get back to xx.

sin⁡(2x)=32,x∈[0,2π] ⟹ sin⁡(u)=32,u∈[0,4π]\sin(2x) = \frac{\sqrt{3}}{2}, \quad x \in [0, 2\pi] \ \Longrightarrow \ \sin(u) = \frac{\sqrt{3}}{2}, \quad u \in [0, 4\pi]

Forget to stretch the domain and you lose the solutions that come from the extra turns. The examiner reports flag this exact failure again and again.

Common ways to lose marks

Even strong students drop marks here in predictable ways. Watch for these.

  • Stopping early. Giving the first solution only, when the period guarantees more. Always sweep the whole domain.
  • Wrong quadrant. Choosing where the function is positive when the equation is negative, or the other way around. The sign decides the quadrants, so check it before placing angles.
  • Mixing up exact values. Swapping π6\frac{\pi}{6} and π3\frac{\pi}{3} is a classic. They are different angles with different values, so keep them straight.
  • Forgetting to stretch the domain for a sin⁡(2x)\sin(2x) or cos⁡(3x)\cos(3x) style equation, which silently halves your answer count.
  • Extra solutions. Listing angles that sit outside the stated domain, or writing a general solution when particular values were asked for. Give only what the domain allows.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

Why does a circular function equation usually have many solutions?
How do you find the reference (base) angle?
What does CAST tell you?
For sin⁡(2x)=a\sin(2x) = a over x∈[0,2π]x \in [0, 2\pi], what is the key step?
What are the periods of sin⁡\sin, cos⁡\cos and tan⁡\tan?
Recall · Solving Exponential and Logarithmic Equations
After substituting and solving, why might you reject a candidate value?
Recall · Functional Notation
What does it mean to solve f(x)=kf(x) = k?

Work through the Worked Examples tab to see the recipe handle cosine, a double angle and tangent, then test yourself in Try It.

Worked examples

Worked Example 1A double angle cosine equation, from a real exam

Let f:[0,2π]→R, f(x)=2cos⁡(2x)+1f : [0, 2\pi] \to R,\ f(x) = 2\cos(2x) + 1. Solve f(x)=0f(x) = 0 for xx.

  1. 1

    Rearrange the equation for cos⁡(2x)\cos(2x).

    2cos⁡(2x)+1=0  ⟹  cos⁡(2x)=−122\cos(2x) + 1 = 0 \implies \cos(2x) = -\frac{1}{2}
  2. 2

    With x∈[0,2π]x\in[0,2\pi], let 2x∈[0,4π]2x\in[0,4\pi] and find all solutions of cos⁡(2x)=−12\cos(2x)=-\tfrac12.

    2x=2π3,4π3,8π3,10π32x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}
  3. 3

    Divide every value by 22. The examiner report notes some students gave only two solutions, ignoring the period.

    x=π3,2π3,4π3,5π3x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}
Answer
x=π3, 2π3, 4π3, 5π3x = \dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ \dfrac{5\pi}{3}

VCAA 2025 Mathematical Methods Exam 1, Q3b

Worked Example 2A sine equation from an integral, from a real exam

Find all values of kk such that ∫0π3sin⁡(x) dx=∫kπ2cos⁡(x) dx\displaystyle\int_0^{\frac{\pi}{3}} \sin(x)\,dx = \int_k^{\frac{\pi}{2}} \cos(x)\,dx, where −3π<k<2π-3\pi < k < 2\pi. (The left integral equals 12\tfrac12.)

  1. 1

    Evaluate the right integral: ∫kπ/2cos⁡(x) dx=[sin⁡x]kπ/2=1−sin⁡(k)\int_k^{\pi/2}\cos(x)\,dx = [\sin x]_k^{\pi/2} = 1 - \sin(k). Set it equal to 12\tfrac12.

    1−sin⁡(k)=12  ⟹  sin⁡(k)=121 - \sin(k) = \frac{1}{2} \implies \sin(k) = \frac{1}{2}
  2. 2

    Solve sin⁡(k)=12\sin(k) = \tfrac12 over −3π<k<2π-3\pi < k < 2\pi using base angle π6\tfrac{\pi}{6}.

    k=π6+2πnork=5π6+2πnk = \frac{\pi}{6} + 2\pi n \quad \text{or} \quad k = \frac{5\pi}{6} + 2\pi n
  3. 3

    Select every solution that lies inside −3π<k<2π-3\pi < k < 2\pi (four of them). Note −5π6-\tfrac{5\pi}{6} is NOT a solution, since sin⁡(−5π6)=−12\sin\left(-\tfrac{5\pi}{6}\right) = -\tfrac12.

    k=−11π6, −7π6, π6, 5π6k = -\frac{11\pi}{6},\ -\frac{7\pi}{6},\ \frac{\pi}{6},\ \frac{5\pi}{6}
Answer
k=−11π6, −7π6, π6, 5π6k = -\dfrac{11\pi}{6},\ -\dfrac{7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}

VCAA 2023 Mathematical Methods Exam 1, Q5b

Worked Example 3A cosine equation

Solve 2cos⁡(x)=−32\cos(x) = -\sqrt{3} for x∈[0,2π]x \in [0, 2\pi].

  1. 1

    Get cos⁡(x)\cos(x) on its own first.

    cos⁡(x)=−32\cos(x) = -\frac{\sqrt{3}}{2}
  2. 2

    Find the reference angle from the exact value, ignoring the sign for now. Here cos⁡(π6)=32\cos\left(\tfrac{\pi}{6}\right) = \tfrac{\sqrt{3}}{2}.

    reference angle=π6\text{reference angle} = \frac{\pi}{6}
  3. 3

    Cosine is negative in the second and third quadrants, so place the reference angle there.

    x=π−π6orx=π+π6x = \pi - \frac{\pi}{6} \quad \text{or} \quad x = \pi + \frac{\pi}{6}
  4. 4

    Work out each value and check both lie inside [0,2π][0, 2\pi].

    x=5π6, 7π6x = \frac{5\pi}{6}, \ \frac{7\pi}{6}
Answer
x=5π6, 7π6x = \dfrac{5\pi}{6}, \ \dfrac{7\pi}{6}
Worked Example 4A double angle, watch the domain

Solve sin⁡(2x)=32\sin(2x) = \dfrac{\sqrt{3}}{2} for x∈[0,2π]x \in [0, 2\pi].

  1. 1

    The angle is 2x2x, not xx. Let u=2xu = 2x and stretch the domain to match. If x∈[0,2π]x \in [0, 2\pi] then u∈[0,4π]u \in [0, 4\pi].

    sin⁡(u)=32,u∈[0,4π]\sin(u) = \frac{\sqrt{3}}{2}, \quad u \in [0, 4\pi]
  2. 2

    Reference angle is π3\tfrac{\pi}{3}. Sine is positive in the first and second quadrants.

    u=π3, 2π3u = \frac{\pi}{3}, \ \frac{2\pi}{3}
  3. 3

    The domain runs to 4π4\pi, so add a full turn of 2π2\pi to each to collect every solution.

    u=π3, 2π3, 7π3, 8π3u = \frac{\pi}{3}, \ \frac{2\pi}{3}, \ \frac{7\pi}{3}, \ \frac{8\pi}{3}
  4. 4

    Now undo the substitution by halving every value, since x=u2x = \tfrac{u}{2}.

    x=π6, π3, 7π6, 4π3x = \frac{\pi}{6}, \ \frac{\pi}{3}, \ \frac{7\pi}{6}, \ \frac{4\pi}{3}
Answer
x=π6, π3, 7π6, 4π3x = \dfrac{\pi}{6}, \ \dfrac{\pi}{3}, \ \dfrac{7\pi}{6}, \ \dfrac{4\pi}{3}
Worked Example 5A tangent equation

Solve tan⁡(x)=−1\tan(x) = -1 for x∈[0,2π]x \in [0, 2\pi].

  1. 1

    Reference angle from the exact value. Here tan⁡(π4)=1\tan\left(\tfrac{\pi}{4}\right) = 1.

    reference angle=π4\text{reference angle} = \frac{\pi}{4}
  2. 2

    Tangent is negative in the second and fourth quadrants.

    x=π−π4orx=2π−π4x = \pi - \frac{\pi}{4} \quad \text{or} \quad x = 2\pi - \frac{\pi}{4}
  3. 3

    Evaluate. Tangent repeats every π\pi, so the two answers sit exactly π\pi apart.

    x=3π4, 7π4x = \frac{3\pi}{4}, \ \frac{7\pi}{4}
Answer
x=3π4, 7π4x = \dfrac{3\pi}{4}, \ \dfrac{7\pi}{4}

Practice questions

Practice test

Try it yourself

8 questions, 11 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.Consider f:R→R, f(x)=2x2+x−1f:R\to R,\ f(x)=2x^{2}+x-1 and g:R→R, g(x)=sin⁡(x)g:R\to R,\ g(x)=\sin(x). The inequality (f∘g)(x)>0(f\circ g)(x)>0 is satisfied when:

1mark
Need a hint?
Let u=sin⁡(x)u = \sin(x) and factor 2u2+u−12u^2 + u - 1, then remember that sin⁡(x)\sin(x) is bounded.

VCAA 2025 Mathematical Methods Exam 2, Section A Q10

Q2.For f(x)=sin⁡(x)+1f(x)=\sin(x)+1 on [0,5π2]\left[0,\dfrac{5\pi}{2}\right], find the exact values of xx for which f(x)=32f(x)=\dfrac{3}{2}.

1mark

Work this on paper. The worked solution appears once you submit.

VCAA 2025 Mathematical Methods Exam 2, Section B Q4b

Q3.For the piecewise function ww with w(t)=h(2t+n)w(t)=h(2t+n) on 20≤t≤27.520\le t\le27.5, where h(t)=−60cos⁡ ⁣(π15t)+75h(t)=-60\cos\!\left(\tfrac{\pi}{15}t\right)+75, find all possible values of nn.

2marks

Work this on paper. The worked solution appears once you submit.

VCAA 2023 Mathematical Methods Exam 2, Section B Q2dii

Q4.The solutions of cos⁡(x)=−12\cos(x) = -\dfrac{1}{2} over x∈[0,2π]x \in [0, 2\pi] are:

1mark
Need a hint?
The reference angle comes from cos⁡(π3)=12\cos\left(\tfrac{\pi}{3}\right) = \tfrac{1}{2}; now ask which two quadrants make cosine negative.

Q5.The solutions of sin⁡(x)=−32\sin(x) = -\dfrac{\sqrt{3}}{2} over x∈[0,2π]x \in [0, 2\pi] are:

1mark
Need a hint?
The reference angle is π3\tfrac{\pi}{3}; a negative sine sits below the axis, so pick the two quadrants where that happens.

Q6.How many solutions does cos⁡(2x)=12\cos(2x) = \dfrac{1}{2} have over x∈[0,2π]x \in [0, 2\pi]?

1mark
Need a hint?
Let u=2xu = 2x and stretch the domain: x∈[0,2π]x \in [0, 2\pi] means uu sweeps two full turns, so count the solutions across all of that.

Q7.The solutions of tan⁡(x)=3\tan(x) = \sqrt{3} over x∈[0,2π]x \in [0, 2\pi] are:

1mark
Need a hint?
The reference angle is π3\tfrac{\pi}{3}, and tangent repeats every π\pi, so the two solutions should sit exactly π\pi apart.

Q8.Solve 2sin⁡(x)−1=02\sin(x) - 1 = 0 for x∈[0,2π]x \in [0, 2\pi]. Give exact values and show your reasoning.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

Why does a trig equation have more than one solution?
Because sine, cosine and tangent repeat. The wave returns to the same height over and over, so within any domain it usually lands on your target value at several different angles, not just one.
How do I find the reference angle?
Ignore the sign for a moment and ask which standard angle gives that value. The exact values for thirty, forty five and sixty degrees, written in radians as pi over six, pi over four and pi over three, are expected knowledge, so learn them by heart.
How do I solve an equation like sin(2x) = a value?
Substitute a single letter for the inside angle and stretch the domain to match. If x runs from zero to two pi, then two x runs from zero to four pi. Solve over that bigger domain, then divide every answer by two to get back to x. Forgetting to stretch the domain is the classic way to lose half the solutions.