Why General Solutions?#
In Grade 10–11, you solved trig equations and gave answers like $\theta = 30°$. But $\sin 30° = \sin 150° = 0.5$. And since trig functions repeat every $360°$, there are actually infinitely many angles that work: $30°$, $150°$, $390°$, $510°$, $-210°$, …
The General Solution is a formula that captures ALL of these angles at once using $n \in \mathbb{Z}$ (where $n$ is any integer).
1. The Three Reference Formulas#
These are the only three patterns you need to memorize. Everything else builds on them.
For $\sin\theta = \sin\alpha$#
$$ \theta = \alpha + n \cdot 360° \quad \text{or} \quad \theta = (180° - \alpha) + n \cdot 360° $$Logic: Sine is positive in Quadrants I and II. The reference angle $\alpha$ gives Q1, and $(180° - \alpha)$ gives Q2. Adding $n \cdot 360°$ accounts for all full rotations.
For $\cos\theta = \cos\alpha$#
$$ \theta = \alpha + n \cdot 360° \quad \text{or} \quad \theta = -\alpha + n \cdot 360° $$Which can be written more compactly as:
$$ \theta = \pm\alpha + n \cdot 360° $$Logic: Cosine is positive in Quadrants I and IV. The reference angle $\alpha$ gives Q1, and $-\alpha$ (or equivalently $360° - \alpha$) gives Q4.
For $\tan\theta = \tan\alpha$#
$$ \theta = \alpha + n \cdot 180° $$Logic: Tangent repeats every $180°$ (not $360°$), so we only need one formula with half the period.
Memory trick: Sine has two formulas with $360°$. Cosine has $\pm$ with $360°$. Tangent has one formula with $180°$.
2. The Strategy for Solving#
Step 1: Simplify the equation#
Use identities (double angle, compound angle, Pythagorean) to reduce the equation so that:
- There is only one trig ratio (all $\sin$, or all $\cos$, or all $\tan$).
- Ideally only one angle (all $\theta$, not a mix of $\theta$ and $2\theta$).
Step 2: Find the reference angle#
Use your calculator or special angles to find $\alpha$ (the reference angle).
Step 3: Write the general solution#
Apply the appropriate formula from Section 1.
Step 4: Find specific solutions (if asked)#
If the question says “for $\theta \in [-180°; 360°]$”, substitute $n = 0, \pm1, \pm2, ...$ into your general solution until you find all values in the given interval.
3. Worked Examples#
Example 1: Basic Sine Equation#
Solve $\sin\theta = 0.5$ for $\theta \in [0°; 360°]$.
Step 1: Already simplified.
Step 2: Reference angle: $\alpha = 30°$ (since $\sin 30° = 0.5$).
Step 3: General solution:
$$ \theta = 30° + n \cdot 360° \quad \text{or} \quad \theta = 150° + n \cdot 360° $$Step 4: For $n = 0$: $\theta = 30°$ or $\theta = 150°$ ✓ For $n = 1$: $\theta = 390°$ (out of range) or $\theta = 510°$ (out of range).
Answer: $\theta = 30°$ or $\theta = 150°$
Example 2: Cosine with Negative Value#
Solve $\cos\theta = -\frac{\sqrt{3}}{2}$ for $\theta \in [-360°; 360°]$.
Step 2: Reference angle: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30°$. But cosine is negative, so the actual angle is in Q2 or Q3: $\alpha = 150°$.
Step 3: General solution:
$$ \theta = \pm 150° + n \cdot 360° $$Step 4:
- $n = 0$: $\theta = 150°$ or $\theta = -150°$ ✓
- $n = 1$: $\theta = 510°$ (out) or $\theta = 210°$ ✓
- $n = -1$: $\theta = -210°$ ✓ or $\theta = -510°$ (out)
Answer: $\theta \in \{-210°; -150°; 150°; 210°\}$
Example 3: Tangent Equation#
Solve $\tan 2\theta = \sqrt{3}$ for $\theta \in [0°; 180°]$.
Step 2: Reference angle: $\tan^{-1}(\sqrt{3}) = 60°$, so $\alpha = 60°$.
Step 3: General solution for $2\theta$:
$$ 2\theta = 60° + n \cdot 180° $$Divide everything by 2 to solve for $\theta$:
$$ \theta = 30° + n \cdot 90° $$Step 4:
- $n = 0$: $\theta = 30°$ ✓
- $n = 1$: $\theta = 120°$ ✓
- $n = 2$: $\theta = 210°$ (out of range)
- $n = -1$: $\theta = -60°$ (out of range)
Answer: $\theta = 30°$ or $\theta = 120°$
Example 4: Using Double Angle Identity First#
Solve $\sin 2\theta - \cos\theta = 0$ for $\theta \in [0°; 360°]$.
Step 1: Expand $\sin 2\theta = 2\sin\theta\cos\theta$:
$$ 2\sin\theta\cos\theta - \cos\theta = 0 $$Factor out $\cos\theta$:
$$ \cos\theta(2\sin\theta - 1) = 0 $$Two equations:
Case 1: $\cos\theta = 0$
$$ \theta = 90° + n \cdot 360° \quad \text{or} \quad \theta = -90° + n \cdot 360° $$In range: $\theta = 90°$ or $\theta = 270°$
Case 2: $2\sin\theta - 1 = 0 \Rightarrow \sin\theta = \frac{1}{2}$
$$ \theta = 30° + n \cdot 360° \quad \text{or} \quad \theta = 150° + n \cdot 360° $$In range: $\theta = 30°$ or $\theta = 150°$
Answer: $\theta \in \{30°; 90°; 150°; 270°\}$
Example 5: Quadratic Trig Equation#
Solve $2\cos^2\theta - \cos\theta - 1 = 0$ for $\theta \in [-180°; 180°]$.
Step 1: This is a quadratic in $\cos\theta$. Let $k = \cos\theta$:
$$ 2k^2 - k - 1 = 0 $$$$ (2k + 1)(k - 1) = 0 $$$$ k = -\frac{1}{2} \quad \text{or} \quad k = 1 $$Case 1: $\cos\theta = -\frac{1}{2}$ Reference: $\cos^{-1}\left(\frac{1}{2}\right) = 60°$, but cosine is negative → $\alpha = 120°$
$$ \theta = \pm 120° + n \cdot 360° $$In range: $\theta = 120°$ or $\theta = -120°$
Case 2: $\cos\theta = 1$
$$ \theta = 0° + n \cdot 360° $$In range: $\theta = 0°$
Answer: $\theta \in \{-120°; 0°; 120°\}$
Example 6: Using $\cos 2\theta$ to Create a Quadratic#
Solve $\cos 2\theta + 3\sin\theta = 2$ for $\theta \in [0°; 360°]$.
Step 1: Replace $\cos 2\theta$ with $1 - 2\sin^2\theta$ (because the equation has $\sin\theta$):
$$ 1 - 2\sin^2\theta + 3\sin\theta = 2 $$$$ -2\sin^2\theta + 3\sin\theta - 1 = 0 $$Multiply by $-1$:
$$ 2\sin^2\theta - 3\sin\theta + 1 = 0 $$Factor:
$$ (2\sin\theta - 1)(\sin\theta - 1) = 0 $$Case 1: $\sin\theta = \frac{1}{2}$
$$ \theta = 30° + n \cdot 360° \quad \text{or} \quad \theta = 150° + n \cdot 360° $$In range: $\theta = 30°$ or $\theta = 150°$
Case 2: $\sin\theta = 1$
$$ \theta = 90° + n \cdot 360° $$In range: $\theta = 90°$
Answer: $\theta \in \{30°; 90°; 150°\}$
4. Special Cases#
When $\sin\theta = 0$#
$$ \theta = n \cdot 180° $$($0°, 180°, 360°, -180°, ...$)
When $\cos\theta = 0$#
$$ \theta = 90° + n \cdot 180° $$($90°, 270°, -90°, ...$)
When $\tan\theta = 0$#
$$ \theta = n \cdot 180° $$(Same as $\sin\theta = 0$, since $\tan\theta = \frac{\sin\theta}{\cos\theta}$)
5. Dealing with Compound Angles in the Unknown#
If the equation is $\sin(\theta + 30°) = \frac{\sqrt{3}}{2}$:
- Treat $(\theta + 30°)$ as a single unit. Let $A = \theta + 30°$.
- Solve $\sin A = \frac{\sqrt{3}}{2}$: $$ A = 60° + n \cdot 360° \quad \text{or} \quad A = 120° + n \cdot 360° $$
- Replace $A$: $$ \theta + 30° = 60° + n \cdot 360° \quad \text{or} \quad \theta + 30° = 120° + n \cdot 360° $$
- Solve for $\theta$: $$ \theta = 30° + n \cdot 360° \quad \text{or} \quad \theta = 90° + n \cdot 360° $$
🚨 Common Mistakes#
- Forgetting the second solution for $\sin$ and $\cos$: $\sin\theta = 0.5$ gives TWO families of solutions ($30°$ and $150°$), not one. Forgetting the second one loses half the marks.
- Wrong period for $\tan$: Tangent repeats every $180°$, not $360°$. If you use $360°$ in the general solution for tangent, you miss half the answers.
- Dividing by a trig function: If you have $\sin\theta\cos\theta = \sin\theta$, do NOT divide both sides by $\sin\theta$. This loses the solutions where $\sin\theta = 0$. Instead, move everything to one side and factor: $\sin\theta(\cos\theta - 1) = 0$.
- Not adjusting for $2\theta$ or $3\theta$: When solving $\sin 2\theta = ...$, the general solution is for $2\theta$. You must divide the entire formula by 2 to get $\theta$.
- Calculator in wrong mode: Your calculator MUST be in degree mode for these questions. Radian mode gives completely different answers.
💡 Pro Tip: The “Number Line” Check#
After finding all solutions in a given interval, plot them on a number line. They should be evenly spaced or follow a clear pattern. If you have solutions at $30°$, $90°$, $150°$, $270°$ — the uneven spacing should prompt you to double-check whether you missed $210°$ or $330°$.
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