How to Solve Trigonometric Equations on Non-Standard Intervals and Write General Solutions
- 4 days ago
- 8 min read
In the first part of this topic, we focused on solving first-degree trigonometric equations on the standard interval:
0° ≤ θ < 360°
or
0 ≤ θ < 2π
That is usually the best place to start because the unit circle is often taught using one full positive rotation.
But not every question gives the standard interval. Sometimes the interval may look like this:
−180° ≤ θ ≤ 180°
or
−π ≤ θ ≤ π
In other cases, the question may not give any restriction at all. When that happens, we need a general solution, which describes all possible coterminal solutions.
This post explains both situations.
Solving Trig Equations on Non-Standard Intervals
A non-standard interval is an interval that does not use the usual one-rotation restriction of:
0° ≤ θ < 360°
or
0 ≤ θ < 2π
For example:
−180° ≤ θ ≤ 180°
is still one full rotation, but it is centered around 0° instead of starting at 0°.
Similarly:
−π ≤ θ ≤ π
is the radian version of the same idea.
The main idea is simple:
Solve the equation on the standard interval first. Then use coterminal angles to rewrite the answers so they fit the given interval.
What Are Coterminal Angles?
Coterminal angles are angles that end in the same position on the unit circle.
In degrees, coterminal angles differ by multiples of 360°.
For example:
240°
and
−120°
are coterminal because:
240° − 360° = −120°
They point in the same direction on the unit circle.
In radians, coterminal angles differ by multiples of 2π.
For example:
4π/3
and
−2π/3
are coterminal because:
4π/3 − 2π = 4π/3 − 6π/3 = −2π/3
So when a question gives a non-standard interval, we are not really finding brand-new angles. We are often rewriting the same unit circle positions using different coterminal representatives.
Method for Non-Standard Intervals
Suppose you are solving a trig equation on an interval like:
−180° ≤ θ ≤ 180°
Use this process:
Step 1: Solve the equation on the standard interval
First solve using:
0° ≤ θ < 360°
This gives the basic unit-circle answers.
Step 2: Check whether those answers fit the given interval
If an answer already fits the interval, keep it.
If it does not fit, add or subtract 360° until it does.
Step 3: Write the answers in the interval requested
The final answers must match the interval from the question, not just the standard interval.
Example 1: Non-Standard Interval in Degrees
Solve:
sinθ = −√3/2, −180° ≤ θ ≤ 180°
This is not the standard interval. The standard interval is usually:
0° ≤ θ < 360°
But the question wants answers between −180° ≤ θ ≤ 180°.
Step 1: Solve on the standard interval first
We are solving:
sinθ = −√3/2
Sine is negative in:
Quadrant III
Quadrant IV
The reference angle is:
60°
because:
sin60° = √3/2
So on the standard interval:
Quadrant III:
180° + 60° = 240°
Quadrant IV:
360° − 60° = 300°
So the standard interval answers are:
240°, 300°
Step 2: Convert these to coterminal angles in the required interval
The required interval is:
−180° ≤ θ ≤ 180°
The angle 240° is too large, so subtract 360°:
240° − 360° = −120°
The angle 300° is also too large, so subtract 360°:
300° − 360° = −60°
Both −120° and −60° fit the interval.
Final answer:
θ = −120°, −60°
Example 2: Non-Standard Interval in Radians
Solve:
cosθ = −1/2, −π ≤ θ ≤ π
The interval:
−π ≤ θ ≤ π
is the radian version of:
−180° ≤ θ ≤ 180°
As usual, we can solve in degrees first and then convert.
Step 1: Solve on the standard interval first
We are solving:
cosθ = −1/2
Cosine is negative in:
Quadrant II
Quadrant III
The reference angle is:
60°
because:
cos60° = 1/2
So on the standard interval:
Quadrant II:
180° − 60° = 120°
Quadrant III:
180° + 60° = 240°
So the standard interval answers are:
120°, 240°
Step 2: Convert to the interval from −180° ≤ θ ≤ 180°
The answer 120° already fits.
The answer 240° is too large, so subtract 360°:
240° − 360° = −120°
So in degrees, the answers are:
−120°, 120°
Step 3: Convert to radians
−120° = −2π/3
120° = 2π/3
Final answer:
θ = −2π/3, 2π/3
Example 3: Calculator-Based Non-Standard Interval
Solve:
tanθ = −0.8, −180° ≤ θ < 180°
This example does not use a special triangle because 0.8 is not one of the standard exact trig values.
Step 1: Determine the correct quadrants
Tangent is negative in:
Quadrant II
Quadrant IV
Step 2: Find the reference angle
Use your calculator:
θᵣ = tan⁻¹(0.8)
θᵣ = 38.7°
Notice that we use positive 0.8 to find the reference angle. The negative sign tells us the quadrants.
Step 3: Solve on the standard interval
In Quadrant II:
180° − 38.7° = 141.3°
In Quadrant IV:
360° − 38.7° = 321.3°
So on the standard interval:
θ = 141.3°, 321.3°
Step 4: Convert to the required interval
The interval is:
−180° ≤ θ < 180°
The angle 141.3° already fits.
The angle 321.3° is too large, so subtract 360°:
321.3° − 360° = −38.7°
Final answer:
θ = −38.7°, 141.3°
Important Note About Endpoints
Always pay close attention to whether the interval includes or excludes its endpoints.
For example:
−180° ≤ θ ≤ 180°
includes both −180° and 180°.
But:
−180° ≤ θ < 180°
includes −180°, but excludes 180°.
This matters because −180° and 180° are coterminal. They point in the same direction, but they are still different numerical angle values.
If the question includes both endpoints, then both can be valid if they satisfy the equation. If the question excludes one endpoint, then you must leave that one out.
Solving Trig Equations with No Restriction
Sometimes a trig equation is given with no interval at all.
For example:
sinθ = 1/2
If the question does not restrict θ, then there are infinitely many solutions.
Why?
Because every full rotation brings us back to the same angle position.
For example:
30°, 390°, 750°, −330°
are all coterminal angles. They all have the same sine value.
So when no restriction is given, we need to write a general solution.
A general solution describes all possible angles that satisfy the equation.
General Solutions in Degrees
In degrees, a full rotation is:
360°
So for sine and cosine equations, we usually take the standard interval answers and add:
360°n
where:
n ∈ ℤ
This means n is any integer:
..., -3, -2, -1, 0, 1, 2, 3, ...
Example 4: General Solution for Sine in Degrees
Solve:
sinθ = 1/2
There is no restriction, so we need a general solution.
Step 1: Solve on the standard interval
Sine is positive in:
Quadrant I
Quadrant II
The reference angle is:
30°
because:
sin30° = 1/2
So on the standard interval:
θ = 30°, 150°
Step 2: Add all coterminal angles
Since sine repeats every 360°, we add:
360°n
to each standard answer.
Final answer:
θ = 30° + 360°n
or
θ = 150° + 360°n
where:
n ∈ ℤ
Example 5: General Solution for Cosine in Degrees
Solve:
cosθ = −√2/2
Step 1: Solve on the standard interval
Cosine is negative in:
Quadrant II
Quadrant III
The reference angle is:
45°
because:
cos45° = √2/2
So the standard interval answers are:
135°, 225°
Step 2: Add all coterminal angles
Cosine repeats every 360°, so we add:
360°n
Final answer:
θ = 135° + 360°n
or
θ = 225° + 360°n
where:
n ∈ ℤ
General Solution for Tangent
Tangent works a little differently.
Sine and cosine repeat every 360°, but tangent repeats every 180°.
That is because tangent has the same value in angles that are 180° apart.
For example:
tan45° = 1
and
tan225° = 1
Those angles differ by 180°:
225° − 45° = 180°
So for tangent equations, we usually only need one answer, then add:
180°n
Example 6: General Solution for Tangent in Degrees
Solve:
tanθ = −√3
Step 1: Find the reference angle
The reference angle is:
60°
because:
tan60° = √3
Step 2: Determine the correct quadrants
Tangent is negative in:
Quadrant II
Quadrant IV
So on the standard interval, the answers would be:
120°, 300°
Step 3: Use the tangent period
Since tangent repeats every 180°, we do not need to write both standard interval answers as separate families.
We can write:
θ = 120° + 180°n
where:
n ∈ ℤ
This works because:
when n = 0:
120° + 180°(0) = 120°
when n = 1:
120° + 180°(1) = 300°
when n = −1:
120° + 180°(−1) = −60°
All of these angles satisfy:
tanθ = −√3
Final answer:
θ = 120° + 180°n, n ∈ ℤ
General Solutions in Radians
In radians, a full rotation is:
2π
So for sine and cosine equations, we usually add:
2πn
For tangent equations, the period is:
π
So for tangent equations, we add:
πn
Example 7: General Solution for Sine in Radians
Solve:
sinθ = −1/2
There is no interval, so we need a general solution.
Step 1: Solve in degrees first
Sine is negative in:
Quadrant III
Quadrant IV
The reference angle is:
30°
So on the standard interval:
θ = 210°, 330°
Step 2: Convert to radians
210° = 7π/6
330° = 11π/6
Step 3: Add the radian period
Since sine repeats every 2π, add 2πn to each answer.
Final answer:
θ = 7π/6 + 2πn
or
θ = 11π/6 + 2πn
where:
n ∈ ℤ
Example 8: General Solution for Tangent in Radians
Solve:
tanθ = 1
Step 1: Solve in degrees first
The reference angle is:
45°
Tangent is positive in:
Quadrant I
Quadrant III
So on the standard interval:
45°, 225°
Step 2: Convert one answer to radians
Since tangent repeats every 180°, we only need one of these answers.
Use:
45° = π/4
Step 3: Add the tangent period
In radians, tangent repeats every:
π
So the general solution is:
θ = π/4 + πn
where:
n ∈ ℤ
Final answer:
θ = π/4 + πn, n ∈ ℤ
Quadrantal General Solutions
Some general solutions involve quadrantal angles.
These are angles lying directly on the axes.
The four basic unit circle points are:
(1, 0), (0, 1), (−1, 0), (0, −1)
They correspond to:
0°, 90°, 180°, 270°,
or:
0, π/2, π, 3π/2
Example 9: General Solution for sinθ = 0
Since sine is the y-coordinate, we want:
y = 0
This happens at:
0°, 180°, 360°, 540°, ...
In other words, it happens every 180°.
So the general solution is:
θ = 180°n, n ∈ ℤ
In radians:
θ = πn, n ∈ ℤ
Example 10: General Solution for cosθ = 0
Since cosine is the x-coordinate, we want:
x = 0
This happens at:
90°, 270°, 450°, ...
These answers are separated by 180°.
So the general solution is:
θ = 90° + 180°n, n ∈ ℤ
In radians:
θ = π/2 + πn, n ∈ ℤ
Summary: Restricted Intervals vs. General Solutions
If the question gives a non-standard interval
For example:
−180° ≤ θ ≤ 180°
or
−π ≤ θ ≤ π
Use this method:
Solve on the standard interval first
Convert the answers to coterminal angles
Keep only the answers that fit the given interval
If the question gives no interval
Then you need a general solution.
For sine and cosine in degrees:
θ = basic answer + 360°n
For sine and cosine in radians:
θ = basic answer + 2πn
For tangent in degrees:
θ = basic answer + 180°n
For tangent in radians:
θ = basic answer + πn
In all cases:
n ∈ ℤ
Final Thoughts
When solving first-degree trigonometric equations, the interval matters.
If the question gives the standard interval, you can solve directly using the unit circle, special triangles, calculator reference angles, or quadrantal angles.
If the question gives a non-standard interval, the easiest strategy is usually to solve on the standard interval first, then rewrite the answers using coterminal angles.
If the question gives no interval at all, then you should not stop at one or two answers. You need a general solution that represents infinitely many coterminal angles.
A strong student should ask three questions before solving:
What trig equation am I solving?
What interval, if any, am I restricted to?
Do I need specific answers or a general solution?
Once those questions are clear, the solving process becomes much more organized.
Need help with solving trigonometric equations on non-standard intervals and writing general solutions? If you’re in Winnipeg and looking for a tutor, Tutor Advance provides expert one-on-one support!