Solving a trigonometric equation means finding every angle in a given interval that makes the equation true. Unlike ordinary algebra, there is usually more than one answer, and missing one is the most common way students lose marks. This guide gives you a reliable method and works through the main equation types you will meet in O/A Level, FSc and university courses.
Why There Are Multiple Solutions
Sine, cosine and tangent are periodic. Sine and cosine repeat every \( 360^\circ \) (or \( 2\pi \)), and tangent repeats every \( 180^\circ \) (or \( \pi \)). So if \( \sin x = 0.5 \), then \( x = 30^\circ \) works, but so does \( 150^\circ \), \( 390^\circ \) and infinitely many more. The question's interval tells you which ones to keep.
The CAST Rule
The CAST diagram tells you which ratios are positive in each quadrant:
- Quadrant 1 (0° to 90°): All are positive.
- Quadrant 2 (90° to 180°): Sine is positive.
- Quadrant 3 (180° to 270°): Tangent is positive.
- Quadrant 4 (270° to 360°): Cosine is positive.
If \( \alpha \) is the acute reference angle (found using the positive value of the ratio), the solutions in \( 0^\circ \le x < 360^\circ \) are:
| Ratio | Positive value | Negative value |
|---|---|---|
| sin | \( \alpha,\ 180^\circ - \alpha \) | \( 180^\circ + \alpha,\ 360^\circ - \alpha \) |
| cos | \( \alpha,\ 360^\circ - \alpha \) | \( 180^\circ - \alpha,\ 180^\circ + \alpha \) |
| tan | \( \alpha,\ 180^\circ + \alpha \) | \( 180^\circ - \alpha,\ 360^\circ - \alpha \) |
The General Method
- Isolate the trigonometric function (or factorise if it is a quadratic).
- Find the reference angle \( \alpha \) using the inverse function on the positive value.
- Use CAST to decide which quadrants the solutions lie in.
- List all solutions in the required interval.
- Check each answer by substituting back.
Worked Example 1: A Linear Equation
Solve \( 2\sin x - 1 = 0 \) for \( 0^\circ \le x < 360^\circ \).
Step 1: Isolate.
$$ 2\sin x = 1 \quad\Rightarrow\quad \sin x = \frac{1}{2} $$
Step 2: Reference angle. \( \alpha = \sin^{-1}\left(\tfrac{1}{2}\right) = 30^\circ \).
Step 3: Quadrants. Sine is positive in quadrants 1 and 2.
Step 4: Solutions.
$$ x = 30^\circ \quad\text{or}\quad x = 180^\circ - 30^\circ = 150^\circ $$
Check: \( \sin 150^\circ = 0.5 \). Both work.
Worked Example 2: A Quadratic Equation
Solve \( 2\cos^2 x - \cos x - 1 = 0 \) for \( 0^\circ \le x < 360^\circ \).
Step 1: Treat it as a quadratic in \( \cos x \). Let \( c = \cos x \):
$$ 2c^2 - c - 1 = 0 \quad\Rightarrow\quad (2c + 1)(c - 1) = 0 $$
Step 2: Solve each factor.
$$ \cos x = -\frac{1}{2} \quad\text{or}\quad \cos x = 1 $$
Step 3: Case \( \cos x = 1 \). This gives \( x = 0^\circ \) (the next one, \( 360^\circ \), is outside the interval).
Step 4: Case \( \cos x = -\tfrac{1}{2} \). Reference angle \( \alpha = \cos^{-1}\left(\tfrac{1}{2}\right) = 60^\circ \). Cosine is negative in quadrants 2 and 3:
$$ x = 180^\circ - 60^\circ = 120^\circ \quad\text{or}\quad x = 180^\circ + 60^\circ = 240^\circ $$
Answer: \( x = 0^\circ,\ 120^\circ,\ 240^\circ \).
Check \( x = 120^\circ \): \( 2\left(-\tfrac12\right)^2 - \left(-\tfrac12\right) - 1 = \tfrac12 + \tfrac12 - 1 = 0 \). Correct.
Worked Example 3: Using an Identity
Solve \( \sin 2x = \cos x \) for \( 0^\circ \le x < 360^\circ \).
Step 1: Use the double angle identity \( \sin 2x = 2\sin x\cos x \):
$$ 2\sin x\cos x = \cos x $$
Step 2: Bring everything to one side and factorise. Do not divide by \( \cos x \), or you will lose solutions.
$$ 2\sin x\cos x - \cos x = 0 \quad\Rightarrow\quad \cos x\,(2\sin x - 1) = 0 $$
Step 3: Solve each factor.
- \( \cos x = 0 \Rightarrow x = 90^\circ,\ 270^\circ \)
- \( \sin x = \tfrac12 \Rightarrow x = 30^\circ,\ 150^\circ \)
Answer: \( x = 30^\circ,\ 90^\circ,\ 150^\circ,\ 270^\circ \).
Worked Example 4: A Multiple Angle
Solve \( \tan 2x = \sqrt{3} \) for \( 0^\circ \le x < 360^\circ \).
Step 1: Adjust the interval. If \( 0^\circ \le x < 360^\circ \), then \( 0^\circ \le 2x < 720^\circ \).
Step 2: Solve for \( 2x \). Reference angle \( \tan^{-1}\sqrt{3} = 60^\circ \). Tangent repeats every \( 180^\circ \):
$$ 2x = 60^\circ,\ 240^\circ,\ 420^\circ,\ 600^\circ $$
Step 3: Divide by 2.
$$ x = 30^\circ,\ 120^\circ,\ 210^\circ,\ 300^\circ $$
Doubling the angle doubles the number of solutions, which is why you must widen the interval first.
General Solutions
University and FSc courses often ask for every solution, written with an integer \( n \):
$$ \sin x = \sin\alpha \Rightarrow x = n\pi + (-1)^n\alpha $$
$$ \cos x = \cos\alpha \Rightarrow x = 2n\pi \pm \alpha $$
$$ \tan x = \tan\alpha \Rightarrow x = n\pi + \alpha $$
For example, \( \tan x = 1 \) gives \( x = n\pi + \dfrac{\pi}{4} \), \( n \in \mathbb{Z} \).
Common Mistakes
- Dividing by a trig function. This loses the solutions where that function equals zero. Factorise instead.
- Forgetting to widen the interval for \( 2x \), \( 3x \) or \( x/2 \).
- Using a negative value to find the reference angle. Always use the positive value, then apply CAST.
- Ignoring impossible values. \( \sin x = 2 \) has no solution, because \( -1 \le \sin x \le 1 \).
- Mixing degrees and radians in the same answer.
Key Takeaways
- Isolate the function, find the reference angle, then use CAST to locate every solution.
- Treat equations like \( 2\cos^2 x - \cos x - 1 = 0 \) as quadratics.
- Use identities to rewrite equations in a single function, then factorise rather than divide.
- For multiple angles, adjust the interval before solving.
- Always check your answers by substituting them back.
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