IB Syllabus Alignment

Standard Level Topic 3.8: Solving trigonometric equations analytically in a given finite interval (both in radians and degrees). Equations of the form $\sin(kx) = c$, $\cos(kx) = c$, $\tan(kx) = c$. Equations requiring the Pythagorean identity or double angle substitutions ($\cos(2x) = 2\cos^2 x - 1$, etc.) leading to quadratics in $\sin x$ or $\cos x$. Factorisation and graphical solutions using technology.

1. Linear Equations & The Altered Domain Protocol

A fundamental skill in IB Mathematics is solving equations where the angle argument is multiplied or shifted, such as $\sin(2x) = c$ or $\cos\left(x - \frac{\pi}{3}\right) = c$. The most common error is failing to adjust the search domain.

  1. Step 1: Adjust the Domain for the Argument: If the original interval for $x$ is $0 \le x \le 2\pi$, and the argument is $kx$: $$0 \le kx \le 2k\pi$$ Example: For $2x$ on $[0, 2\pi]$, you must find all solutions for $2x$ in $[0, 4\pi]$ (two full revolutions of the unit circle!).
  2. Step 2: Find the Reference Angle ($\alpha$): Find the acute angle $\alpha \in \left[0, \frac{\pi}{2}\right]$ satisfying the positive ratio: $$\alpha = \arcsin(|c|) \qquad \text{or} \qquad \alpha = \arccos(|c|)$$
  3. Step 3: Determine the Quadrants via ASTC: - If $c > 0$, select quadrants where the ratio is positive. - If $c < 0$, select quadrants where the ratio is negative.
  4. Step 4: Generate All Values in the Altered Domain: Add full rotations of $+2\pi$ (or $+360^\circ$) to cycle 1 solutions until the upper boundary of the altered domain is reached.
  5. Step 5: Solve for $x$: Divide every candidate angle by $k$ (or reverse the phase shift) to obtain the final values of $x$.
Worked Example 1.1 — Solving with Multiple Frequency

Find all exact solutions to the equation $2\sin(2x) = -\sqrt{3}$ for $0 \le x \le 2\pi$.

  1. Step 1: Isolate the trigonometric function: $$\sin(2x) = -\frac{\sqrt{3}}{2}$$
  2. Step 2: Adjust domain: Since $0 \le x \le 2\pi$, the argument satisfies $0 \le 2x \le 4\pi$.
  3. Step 3: Find reference angle $\alpha$: $$\sin\alpha = \frac{\sqrt{3}}{2} \implies \alpha = \frac{\pi}{3}$$
  4. Step 4: Find solutions in $[0, 4\pi]$: Sine is negative in Quadrants III and IV. - First Revolution $[0, 2\pi]$: $$2x_1 = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$$ $$2x_2 = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$$ - Second Revolution $[2\pi, 4\pi]$: Add $2\pi = \frac{6\pi}{3}$: $$2x_3 = \frac{4\pi}{3} + 2\pi = \frac{10\pi}{3}$$ $$2x_4 = \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3}$$
  5. Step 5: Divide by 2 to solve for $x$: $$x = \frac{4\pi}{6} = \frac{2\pi}{3}, \quad x = \frac{5\pi}{6}, \quad x = \frac{10\pi}{6} = \frac{5\pi}{3}, \quad x = \frac{11\pi}{6}$$
Answer: $x = \frac{2\pi}{3}, \; \frac{5\pi}{6}, \; \frac{5\pi}{3}, \; \frac{11\pi}{6}$.

2. Quadratic Trigonometric Equations

Equations with squared trigonometric terms require factoring or the quadratic formula. If the equation mixes sine and cosine, the Pythagorean identity must be used to express the equation in terms of a single trigonometric ratio.

  • Single Variable Quadratic Form: $$a\sin^2 x + b\sin x + c = 0 \quad \text{or} \quad a\cos^2 x + b\cos x + c = 0$$ Let $u = \sin x$ (or $u = \cos x$) to solve $au^2 + bu + c = 0$.
  • Eliminating Mixed Functions: If an equation contains $\cos^2 x$ and $\sin x$, replace $\cos^2 x$ with $1 - \sin^2 x$: $$2\cos^2 x + 3\sin x - 3 = 0 \implies 2(1 - \sin^2 x) + 3\sin x - 3 = 0 \implies -2\sin^2 x + 3\sin x - 1 = 0$$
  • Critical Boundary Check: Because real sine and cosine functions are bounded between $-1$ and $+1$: $$-1 \le \sin x \le 1 \qquad \text{and} \qquad -1 \le \cos x \le 1$$ Any factor yielding $\sin x = 2$ or $\cos x = -1.5$ has no real solutions and must be explicitly rejected with a note.
Worked Example 1.2 — Quadratic Trigonometric Factorisation

Solve the equation $2\cos^2 x + \sin x - 1 = 0$ for $0 \le x \le 2\pi$.

  1. Step 1: Convert to a single function using $\cos^2 x = 1 - \sin^2 x$: $$2(1 - \sin^2 x) + \sin x - 1 = 0$$ $$2 - 2\sin^2 x + \sin x - 1 = 0$$ $$-2\sin^2 x + \sin x + 1 = 0 \implies 2\sin^2 x - \sin x - 1 = 0$$
  2. Step 2: Factor the quadratic: $$(2\sin x + 1)(\sin x - 1) = 0$$ This produces two separate linear cases: $$\sin x = -\frac{1}{2} \qquad \text{or} \qquad \sin x = 1$$
  3. Step 3: Solve each branch in $[0, 2\pi]$: - Branch 1: $\sin x = -\frac{1}{2}$
    Reference angle $\alpha = \frac{\pi}{6}$, Quadrants III and IV: $$x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}, \qquad x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$$ - Branch 2: $\sin x = 1$
    Occurs at the top of the unit circle: $$x = \frac{\pi}{2}$$
Answer: $x = \frac{\pi}{2}, \; \frac{7\pi}{6}, \; \frac{11\pi}{6}$.

Visualizing Trigonometric Solutions Graphically

The diagram below illustrates solving $\sin(2x) = \frac{\sqrt{3}}{2}$ over $[0, 2\pi]$, showing the 4 points of intersection between the wave and the horizontal line.

x y O π/2 π 3π/2 2π y = √3/2 π/6 π/3 7π/6 4π/3 Equation: sin(2x) = √3/2 4 Real Solutions in [0, 2π]
Figure 3.6 — Graphical solution of $\sin(2x) = \frac{\sqrt{3}}{2}$ over $0 \le x \le 2\pi$, illustrating the 4 distinct intersection roots produced by two full periods of the wave.

3. Double Angle Equations & Factorisation Warnings

Equations combining single and double angles (e.g. $\sin(2x) = \cos x$ or $\cos(2x) + \cos x = 0$) require identity substitutions followed by careful factorisation.

  • The "Never Divide by a Variable" Rule: Consider the equation $\sin(2x) = \cos x$. Expanding gives: $$2\sin x\cos x = \cos x$$ CRITICAL WARNING: If you divide both sides by $\cos x$, you get $2\sin x = 1 \implies \sin x = \frac{1}{2}$, losing the solutions where $\cos x = 0$!
    Instead, bring everything to one side and factorise: $$2\sin x\cos x - \cos x = 0 \implies \cos x(2\sin x - 1) = 0$$ $$\cos x = 0 \qquad \text{or} \qquad \sin x = \frac{1}{2}$$
  • Double Angle Cosine Substitution in Equations: When solving an equation mixing $\cos(2x)$ and $\cos x$, replace $\cos(2x)$ with $2\cos^2 x - 1$ to obtain a quadratic in $\cos x$: $$\cos(2x) + \cos x = 0 \implies (2\cos^2 x - 1) + \cos x = 0 \implies 2\cos^2 x + \cos x - 1 = 0$$ When mixing $\cos(2x)$ and $\sin x$, replace $\cos(2x)$ with $1 - 2\sin^2 x$.
Worked Example 1.3 — Solving with Double Angle Cosine

Find all exact solutions to the equation $\cos(2x) = \sin x$ for $0 \le x \le 2\pi$.

  1. Step 1: Express entirely in terms of $\sin x$: Substitute $\cos(2x) = 1 - 2\sin^2 x$: $$1 - 2\sin^2 x = \sin x \implies 2\sin^2 x + \sin x - 1 = 0$$
  2. Step 2: Factor the quadratic: $$(2\sin x - 1)(\sin x + 1) = 0$$ $$\sin x = \frac{1}{2} \qquad \text{or} \qquad \sin x = -1$$
  3. Step 3: Solve each branch in $[0, 2\pi]$: - For $\sin x = \frac{1}{2}$ (Quadrants I and II): $$x = \frac{\pi}{6}, \qquad x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ - For $\sin x = -1$ (Bottom of unit circle): $$x = \frac{3\pi}{2}$$
Answer: $x = \frac{\pi}{6}, \; \frac{5\pi}{6}, \; \frac{3\pi}{2}$.

4. Factoring Double Angles & Scaled Arguments

Worked Example 1.4 — Solving $\sin(2x) = \cos x$ via Factoring

Solve the equation $\sin(2x) = \cos x$ for $0 \le x \le 2\pi$. [5 marks]

  1. Step 1: Apply double angle identity $\sin(2x) = 2\sin x\cos x$: $$2\sin x\cos x = \cos x \quad [M1]$$
  2. Step 2: Collect all terms on one side and factorise (DO NOT divide by $\cos x$!): $$2\sin x\cos x - \cos x = 0 \implies \cos x(2\sin x - 1) = 0 \quad [M1]$$
  3. Step 3: Branch 1 ($\cos x = 0$): $$x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2} \quad [A1]$$
  4. Step 4: Branch 2 ($2\sin x - 1 = 0 \implies \sin x = \frac{1}{2}$): $$\text{In Quadrants I and II: } x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6} \quad [A1]$$
  5. Step 5: Assemble all solutions in ascending order: $$x = \frac{\pi}{6}, \quad \frac{\pi}{2}, \quad \frac{5\pi}{6}, \quad \frac{3\pi}{2} \quad [A1]$$
Answers: $x \in \left\{ \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2} \right\}$.
Worked Example 1.5 — Solving with an Adjusted Argument

Solve the equation $2\sin\left(2x - \frac{\pi}{3} \right) = \sqrt{3}$ for $0 \le x \le \pi$. [5 marks]

  1. Step 1: Isolate the sine expression: $$\sin\left(2x - \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \quad [M1]$$
  2. Step 2: Determine the domain for the compound argument $\theta = 2x - \frac{\pi}{3}$: $$0 \le x \le \pi \implies 0 \le 2x \le 2\pi \implies -\frac{\pi}{3} \le 2x - \frac{\pi}{3} \le \frac{5\pi}{3} \quad [M1]$$
  3. Step 3: Find all angles where $\sin\theta = \frac{\sqrt{3}}{2}$ in this interval: $$\theta_1 = \frac{\pi}{3}, \quad \theta_2 = \frac{2\pi}{3} \quad [A1]$$
  4. Step 4: Solve for $x$ in each case: $$2x - \frac{\pi}{3} = \frac{\pi}{3} \implies 2x = \frac{2\pi}{3} \implies x = \frac{\pi}{3} \quad [A1]$$ $$2x - \frac{\pi}{3} = \frac{2\pi}{3} \implies 2x = \pi \implies x = \frac{\pi}{2} \quad [A1]$$
Answers: $x = \frac{\pi}{3}$ or $x = \frac{\pi}{2}$.

10 Tiered Practice Exercises

Progress from procedural linear equations in Tier 1 to quadratic trigonometric forms and double angle substitutions in Tier 2, concluding with boundary analyses in Tier 3.

Tier 1: Core Fluency (Problems 1–4)

Exercise 1 Linear Exact Equation

Solve $2\cos x + \sqrt{3} = 0$ for $0 \le x \le 2\pi$, giving exact answers.

View Full Solution

Step 1: Isolate $\cos x$:
$$\cos x = -\frac{\sqrt{3}}{2}$$

Step 2: Reference angle & quadrants:
$\cos\alpha = \frac{\sqrt{3}}{2} \implies \alpha = \frac{\pi}{6}$.
Cosine is negative in Quadrants II and III: $$x_1 = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ $$x_2 = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$$

Final Answer: $x = \frac{5\pi}{6}, \frac{7\pi}{6}$.

Exercise 2 Altered Domain with Multiple Frequency

Solve $\sin(3x) = 1$ for $0 \le x \le \pi$.

View Full Solution

Step 1: Adjust domain:
$0 \le x \le \pi \implies 0 \le 3x \le 3\pi$.

Step 2: Solve for $3x$:
$\sin\theta = 1$ at $\theta = \frac{\pi}{2}, \frac{5\pi}{2}$ inside $[0, 3\pi]$: $$3x = \frac{\pi}{2} \implies x = \frac{\pi}{6}$$ $$3x = \frac{5\pi}{2} \implies x = \frac{5\pi}{6}$$

Final Answer: $x = \frac{\pi}{6}, \frac{5\pi}{6}$.

Exercise 3 Equations Reducible to Tangent

Solve $3\sin\theta - \sqrt{3}\cos\theta = 0$ for $0^\circ \le \theta \le 360^\circ$.

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Step 1: Divide by $\cos\theta$ (since $\cos\theta = 0$ is not a solution):
$$3\frac{\sin\theta}{\cos\theta} = \sqrt{3} \implies 3\tan\theta = \sqrt{3} \implies \tan\theta = \frac{\sqrt{3}}{3}$$

Step 2: Solve in degrees:
Reference angle $\alpha = 30^\circ$. Tangent is positive in Quadrants I and III: $$\theta_1 = 30^\circ$$ $$\theta_2 = 180^\circ + 30^\circ = 210^\circ$$

Final Answer: $\theta = 30^\circ, 210^\circ$.

Exercise 4 Factoring Common Terms

Solve $\sin(2x) - \sin x = 0$ for $0 \le x \le 2\pi$.

View Full Solution

Step 1: Use double angle identity:
$$2\sin x\cos x - \sin x = 0$$

Step 2: Factor out $\sin x$:
$$\sin x(2\cos x - 1) = 0$$ $$\sin x = 0 \qquad \text{or} \qquad \cos x = \frac{1}{2}$$

Step 3: Solve each factor:
- For $\sin x = 0$: $x = 0, \pi, 2\pi$.
- For $\cos x = \frac{1}{2}$: $x = \frac{\pi}{3}, 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.

Final Answer: $x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi$.

Tier 2: Exam Modeling (Problems 5–8)

Exercise 5 Quadratic Trigonometric Equation

Solve the equation $2\sin^2 x - 3\sin x + 1 = 0$ for $0 \le x \le 2\pi$.

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Step 1: Factor the quadratic:
$$(2\sin x - 1)(\sin x - 1) = 0$$ $$\sin x = \frac{1}{2} \qquad \text{or} \qquad \sin x = 1$$

Step 2: Solve in $[0, 2\pi]$:
- For $\sin x = \frac{1}{2}$: $x = \frac{\pi}{6}, \frac{5\pi}{6}$.
- For $\sin x = 1$: $x = \frac{\pi}{2}$.

Final Answer: $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$.

Exercise 6 Double Angle Cosine Quadratic

Solve $\cos(2x) + 3\cos x - 1 = 0$ for $0 \le x \le 2\pi$, giving answers to 3 significant figures.

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Step 1: Substitute $\cos(2x) = 2\cos^2 x - 1$:
$$(2\cos^2 x - 1) + 3\cos x - 1 = 0 \implies 2\cos^2 x + 3\cos x - 2 = 0$$

Step 2: Factor:
$$(2\cos x - 1)(\cos x + 2) = 0$$ $$\cos x = \frac{1}{2} \qquad \text{or} \qquad \cos x = -2$$

Step 3: Boundary check & solve:
- $\cos x = -2$ has no real solutions since $-1 \le \cos x \le 1$.
- $\cos x = \frac{1}{2} \implies x = \frac{\pi}{3} \approx 1.05\text{ rad}, \quad x = \frac{5\pi}{3} \approx 5.24\text{ rad}$.

Final Answer: $x = \frac{\pi}{3}, \frac{5\pi}{3}$ (or $1.05, 5.24$).

Exercise 7 Phase Shift Equation

Solve $2\cos\left(x - \frac{\pi}{4}\right) = 1$ for $0 \le x \le 2\pi$.

View Full Solution

Step 1: Isolate cosine:
$$\cos\left(x - \frac{\pi}{4}\right) = \frac{1}{2}$$

Step 2: Adjust domain:
$-\frac{\pi}{4} \le x - \frac{\pi}{4} \le \frac{7\pi}{4}$.

Step 3: Solve for argument:
Cosine equals $\frac{1}{2}$ at $\pm\frac{\pi}{3}$: $$x - \frac{\pi}{4} = \frac{\pi}{3} \implies x = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12}$$ $$x - \frac{\pi}{4} = -\frac{\pi}{3} \implies x = -\frac{\pi}{3} + \frac{\pi}{4} = -\frac{\pi}{12} \quad (\text{outside }[0, 2\pi])$$ Next positive angle: $2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$: $$x - \frac{\pi}{4} = \frac{5\pi}{3} \implies x = \frac{5\pi}{3} + \frac{\pi}{4} = \frac{20\pi + 3\pi}{12} = \frac{23\pi}{12}$$

Final Answer: $x = \frac{7\pi}{12}, \frac{23\pi}{12}$.

Exercise 8 Squaring and Extraneous Roots

Solve $\cos x + \sin x = 1$ for $0 \le x < 2\pi$.

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Method 1: Square both sides:
$$(\cos x + \sin x)^2 = 1^2 \implies \cos^2 x + 2\sin x\cos x + \sin^2 x = 1$$ $$1 + \sin(2x) = 1 \implies \sin(2x) = 0$$ For $x \in [0, 2\pi)$, $2x \in [0, 4\pi)$: $$2x = 0, \pi, 2\pi, 3\pi \implies x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$$

Step 2: Check for extraneous roots introduced by squaring:
- $x = 0$: $\cos 0 + \sin 0 = 1 + 0 = 1$. (Valid)
- $x = \frac{\pi}{2}$: $\cos(\pi/2) + \sin(\pi/2) = 0 + 1 = 1$. (Valid)
- $x = \pi$: $\cos\pi + \sin\pi = -1 + 0 = -1 \neq 1$. (Extraneous, reject)
- $x = \frac{3\pi}{2}$: $\cos(3\pi/2) + \sin(3\pi/2) = 0 + (-1) = -1 \neq 1$. (Extraneous, reject)

Final Answer: $x = 0, \frac{\pi}{2}$.

Tier 3: Synthesis & Problem-Solving (Problems 9–10)

Exercise 9 Multi-Root Polynomial Sum

Find all exact solutions to $4\sin^3 x - 3\sin x = 0$ for $0 \le x \le 2\pi$, and calculate the sum of all distinct solutions.

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Step 1: Factor out $\sin x$:
$$\sin x(4\sin^2 x - 3) = 0$$ $$\sin x = 0 \qquad \text{or} \qquad \sin^2 x = \frac{3}{4} \implies \sin x = \pm\frac{\sqrt{3}}{2}$$

Step 2: Collect all solutions in $[0, 2\pi]$:
- From $\sin x = 0$: $x = 0, \pi, 2\pi$.
- From $\sin x = \frac{\sqrt{3}}{2}$: $x = \frac{\pi}{3}, \frac{2\pi}{3}$.
- From $\sin x = -\frac{\sqrt{3}}{2}$: $x = \frac{4\pi}{3}, \frac{5\pi}{3}$.

Step 3: Sum the distinct solutions:
$$\text{Sum} = 0 + \frac{\pi}{3} + \frac{2\pi}{3} + \pi + \frac{4\pi}{3} + \frac{5\pi}{3} + 2\pi$$ $$\text{Sum} = \frac{(0 + 1 + 2 + 3 + 4 + 5 + 6)\pi}{3} = \frac{21\pi}{3} = 7\pi$$

Final Answers: $x = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$; Sum $= 7\pi$.

Exercise 10 Parametric Existence Condition

Consider the equation $\cos^2 x - 2\cos x + k = 0$.
(a) Find the set of real values of $k$ for which the equation has at least one real solution.
(b) Solve the equation completely for $0 \le x \le 2\pi$ when $k = -3$.

View Full Solution

(a) Condition on $k$:
Let $u = \cos x$, with $-1 \le u \le 1$.
$$u^2 - 2u + k = 0 \implies k = 2u - u^2$$ Let $g(u) = 2u - u^2$ on $u \in [-1, 1]$:
- Maximum occurs at vertex $u = -\frac{b}{2a} = 1$: $g(1) = 2(1) - 1^2 = 1$.
- Minimum occurs at endpoint $u = -1$: $g(-1) = 2(-1) - (-1)^2 = -2 - 1 = -3$.
Thus, for a solution $u \in [-1, 1]$ to exist: $$-3 \le k \le 1$$

(b) Solve when $k = -3$:
$$u^2 - 2u - 3 = 0 \implies (u - 3)(u + 1) = 0$$ - $u = 3 \implies \cos x = 3$ (no real solution).
- $u = -1 \implies \cos x = -1 \implies x = \pi$.

Final Answers: (a) $-3 \le k \le 1$; (b) $x = \pi$.

Solving Equations Reference Sheet

In the IB Formula Booklet (Topic 3.6 & 3.8)

The official IB Mathematics AA SL formula booklet lists the essential identities for equation solving:

Pythagorean Identity $$\cos^2\theta + \sin^2\theta = 1$$
Tangent Definition $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$
Double Angle Sine $$\sin(2\theta) = 2\sin\theta\cos\theta$$
Double Angle Cosine $$\cos(2\theta) = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$$
Must Memorize & Strategic Checklists
  • Altered Domain Search: Always multiply the boundary by $k$ before solving: $$x \in [0, 2\pi] \implies kx \in [0, 2k\pi]$$
  • Quadratic Conversion Protocol: - If equation contains $\sin^2 x$ and $\cos x$, replace $\sin^2 x$ with $1 - \cos^2 x$. - If equation contains $\cos(2x)$ and $\cos x$, replace $\cos(2x)$ with $2\cos^2 x - 1$. - If equation contains $\cos(2x)$ and $\sin x$, replace $\cos(2x)$ with $1 - 2\sin^2 x$.
  • Boundaries of Sine & Cosine: $|\sin x| \le 1$ and $|\cos x| \le 1$. Automatically reject roots $|u| > 1$.
Critical IB Examination Pitfalls
  • Pitfall 1: Dividing by a Common Trigonometric Variable: In $\sin(2x) = \sin x$, dividing through by $\sin x$ discards all solutions where $\sin x = 0$ ($x = 0, \pi, 2\pi$), forfeiting the majority of marks.
  • Pitfall 2: Forgetting Extra Rotations for Multiple Angles: Searching only up to $2\pi$ for $2x$, producing only 2 answers instead of 4.
  • Pitfall 3: Not Checking Endpoints: Failing to note whether the domain is open ($<$) or closed ($\le$). For instance, if interval is $0 \le x < 2\pi$, $2\pi$ must be excluded!
  • Pitfall 4: GDC Degree Mode vs Radian Mode: Leaving GDC in Degree mode when solving on Paper 2 for radian bounds produces completely erroneous answers.

Section A: Diagnostic Multiple Choice

Answer the questions to check your understanding and score.

Score: 0 / 4

Question 1: How many real solutions does the equation $\sin(3x) = 0.5$ have in the interval $0 \le x \le 2\pi$?

Question 2: What is the solution to $\cos x = -\frac{\sqrt{2}}{2}$ in the restricted interval $0 \le x \le \pi$?

Question 3: Which identity substitution should be used to solve $2\cos(2x) + \sin x - 1 = 0$ as a quadratic in a single function?

Question 4: What are the solutions to $2\cos x\sin x = \cos x$ in the interval $0 \le x \le \pi$?

Section B: Exam-Style Free-Response Questions

Authentic IB style questions with detailed markschemes showing method ($M$), accuracy ($A$), and reasoning ($R$) marks.

Question 5 (Paper 1 Style) 6 Marks

Consider the equation $2\sin(2x) = \tan x$ for $0 \le x \le \pi$.
(a) Show that the equation can be written in the form $\sin x(4\cos^2 x - 1) = 0$. [2]
(b) Hence find all exact solutions in the interval $[0, \pi]$. [4]

View Full Markscheme

(a) Show algebraic form: [2 marks]
Substitute $\sin(2x) = 2\sin x\cos x$ and $\tan x = \frac{\sin x}{\cos x}$: $$2(2\sin x\cos x) = \frac{\sin x}{\cos x} \implies 4\sin x\cos x = \frac{\sin x}{\cos x} \quad (M1)$$ Multiply by $\cos x$ (for $\cos x \neq 0$): $$4\sin x\cos^2 x - \sin x = 0 \implies \sin x(4\cos^2 x - 1) = 0 \quad (A1)$$

(b) Find all exact solutions: [4 marks]
From $\sin x = 0$: $$x = 0, \pi \quad (A1)$$ From $4\cos^2 x - 1 = 0 \implies \cos^2 x = \frac{1}{4} \implies \cos x = \pm\frac{1}{2}$ (M1):
- For $\cos x = \frac{1}{2}$: $x = \frac{\pi}{3}$ (A1)
- For $\cos x = -\frac{1}{2}$: $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ (A1)
Complete solution set: $$x = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi$$

Question 6 (Paper 2 Style) 8 Marks

The functions $f$ and $g$ are defined by $f(x) = 3\cos(2x) + 1$ and $g(x) = 2\sin x$ for $0 \le x \le 2\pi$.
(a) Show that the $x$-coordinates of the intersections of $y = f(x)$ and $y = g(x)$ satisfy the equation $6\sin^2 x + 2\sin x - 4 = 0$. [2]
(b) Solve this quadratic equation to find the exact values of $\sin x$. [2]
(c) Hence determine the $x$-coordinates of all intersection points in the interval $[0, 2\pi]$, giving exact values where possible and other values to 3 significant figures. [4]

View Full Markscheme

(a) Show quadratic equation: [2 marks]
Equate functions: $$3\cos(2x) + 1 = 2\sin x \quad (M1)$$ Substitute $\cos(2x) = 1 - 2\sin^2 x$: $$3(1 - 2\sin^2 x) + 1 = 2\sin x \implies 3 - 6\sin^2 x + 1 = 2\sin x$$ $$4 - 6\sin^2 x = 2\sin x \implies 6\sin^2 x + 2\sin x - 4 = 0 \quad (A1)$$

(b) Solve quadratic in $\sin x$: [2 marks]
Divide by 2: $$3\sin^2 x + \sin x - 2 = 0 \implies (3\sin x - 2)(\sin x + 1) = 0 \quad (M1)$$ $$\sin x = \frac{2}{3} \qquad \text{or} \qquad \sin x = -1 \quad (A1)$$

(c) Find $x$-coordinates: [4 marks]
- Case 1: $\sin x = \frac{2}{3}$
Reference angle $\alpha = \arcsin(2/3) \approx 0.7297\text{ rad}$ (M1)
Quadrant I: $x_1 \approx 0.730\text{ rad}$ (A1)
Quadrant II: $x_2 = \pi - 0.7297 \approx 2.41\text{ rad}$ (A1)
- Case 2: $\sin x = -1$
$x_3 = \frac{3\pi}{2}$ (exact) (A1)
The $x$-coordinates are $x \approx 0.730, \; 2.41, \; \frac{3\pi}{2}$.