01 · Positive secant
Solve sec x = 2 for 0 ≤ x < 2π.
Hint
Cos x = 1/2.
Worked solution
x = π/3, 5π/3. Both make the original denominator nonzero.
Understand · explore · practise
Sec, cosec and cot equations
Solve sec, cosec and cot equations in degrees and radians. Find every interval solution, handle shifted angles and factors, and reject excluded inputs.
Before you startReciprocal trig definitions and graphs; solving sine, cosine and tangent equations
01 / Rewrite the equation and keep the domain
Use a familiar trig equation, then check the original one.
For sec or cosec equal to a nonzero constant, take reciprocals to get a cosine or sine equation. For cot, use cos/sin, with a special check when the target is 0.
sec u = k → cos u = 1/k
cosec u = k → sin u = 1/k
For these two, |k| ≥ 1 is required.
There are no real sec or cosec solutions when |k| < 1, including k = 0. Cot can equal any real number. Use the graph to predict the number of roots, then solve algebraically.
Find every intersection in the intervalExplore
2 solutions: π/3, 5π/3.
Sec and cosec need |k| ≥ 1. Cot can take any real k.
Both interval endpoints are included when the function is defined. Decimal angles are approximations in radians. Dashed vertical lines are excluded inputs, never solutions.
Watch a cotangent root pass through zero
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Solve a secant equation
Cosine gives two branches in a typical full period.
Solve sec x = −2 on −π ≤ x ≤ πWorked example
cos x = −1/2
The original requires cosine nonzero; these candidates satisfy that.
Reference angle = π/3
Cosine is negative in Quadrants II and III.
x = −2π/3 or 2π/3
These are the two angles in the stated interval.
Check 1/(−1/2) = −2
Neither is an excluded input.
For sec x = 1 on 0 ≤ x ≤ 2π, the solutions are 0 and 2π. They are different allowed endpoints even though they share a terminal ray. If the upper endpoint is excluded, keep only 0.
03 / Solve a cosecant equation
Use sine’s signs and retain the full interval.
Solve cosec x = −2 on 0 ≤ x < 2πWorked example
sin x = −1/2
The reference angle is π/6.
Sine is negative in Quadrants III and IV
Use π +π/6 and 2π −π/6.
x = 7π/6 or 11π/6
Both give cosec x = −2.
At the range edge, cosec x = 1 gives only x = π/2 in this interval. Do not create a second distinct angle by applying two sine branches that coincide.
04 / Treat cot x = 0 directly
The reciprocal shortcut needs a nonzero target.
cot u = k, k ≠ 0 → tan u = 1/k
cot u = 0 → cos u = 0, with sin u ≠ 0
Solve cot x = 0 on −π ≤ x ≤ πWorked example
cos x/sin x = 0
A quotient is zero when its numerator is zero and denominator nonzero.
x = −π/2 or π/2
At both angles sine is ±1.
Do not write tan x = 1/0
That would discard the valid zero-cotangent solutions.
Solve cot x = −1 on 0 ≤ x < 2πWorked example
tan x = −1
The target is nonzero, so this reciprocal step is valid.
x = 3π/4 or 7π/4
Tangent and cotangent repeat after π.
05 / Find numerical roots in the requested unit
A principal calculator angle is only the starting point.
Solve 3cot x + 2 = 0 for 0° ≤ x < 360°Worked example
cot x = −2/3, so tan x = −3/2
Use DEG mode.
arctan(−3/2) ≈ −56.3099°
This principal angle is outside the requested interval.
Add 180° repeatedly: x ≈ 123.6901°, 303.6901°
Tangent’s period is 180°.
To 1 decimal place: x = 123.7°, 303.7°
Round only after generating all solutions.
In radians the tangent period is π. Always check the displayed calculator mode, the interval endpoints and the accuracy requested in the question.
06 / Transform the entire interval
Solve for the inner angle before mapping back.
Solve sec(2x − π/6) = 2 for 0 ≤ x ≤ 2πWorked example
Let u = 2x −π/6; then −π/6 ≤ u ≤ 23π/6
Both interval endpoints must be transformed.
cos u = 1/2
All roots in this inner interval are needed.
u = π/3, 5π/3, 7π/3, 11π/3
These are the two cosine branches repeated over the interval.
x = (u +π/6)/2
Undo the shift and scale.
x = π/4, 11π/12, 5π/4, 23π/12
Each lies in the original interval and makes sec equal 2.
Solve cosec(x + π/4) = −√2 for 0 ≤ x < 2πWorked example
sin(x +π/4) = −√2/2
The inner interval is π/4 ≤ u < 9π/4.
u = 5π/4 or 7π/4
Subtract π/4 after finding both sine branches.
x = π or 3π/2
Check in the original equation.
07 / Use each factor instead of dividing it away
A zero factor may supply valid roots.
Solve cot x(2sin x −1) = 0 for 0 ≤ x < 2πWorked example
Original domain: sin x ≠ 0
Cot must exist.
Either cot x = 0 or sin x = 1/2
The product can be zero through either factor.
cot x = 0 gives x = π/2, 3π/2
These roots disappear if you divide by cot x.
sin x = 1/2 gives x = π/6, 5π/6
These also satisfy the original domain.
Complete set: π/6, π/2, 5π/6, 3π/2
Sort and remove only duplicates, not whole branches.
Before dividing by a trig factor, consider its zero case separately. An algebraically shorter equation may no longer be equivalent to the original.
08 / Check candidates after clearing denominators
A derived equation can include an input the original never allowed.
Solve sec x = tan x for 0 ≤ x < 2πWorked example
The original needs cos x ≠ 0
Both functions divide by cosine.
Multiplying by cos x gives 1 = sin x
This step is valid only on the original domain.
The derived candidate is x = π/2
But cosine is 0 there.
Reject it: there are no solutions
The original equation is undefined at the candidate.
Solve sec x = 2tan x on the same intervalWorked example
With cos x ≠ 0, the equation becomes 1 = 2sin x
Clear the common denominator.
x = π/6 or 5π/6
Cosine is nonzero at both, so both are valid.
A closed interval does not override the function’s domain. An included numerical boundary still cannot be a solution if the original expression is undefined there.
09 / Keep both signs when taking square roots
A squared equation can produce four angles per full turn.
Solve sec²x = 4 for 0 ≤ x < 2πWorked example
sec x = 2 or sec x = −2
Both signs square to 4.
cos x = 1/2 or cos x = −1/2
Each target is allowed.
x = π/3, 2π/3, 4π/3, 5π/3
There are four valid roots, all with nonzero cosine.
More complicated equations involving squared reciprocal functions can be turned into algebraic quadratics using the sec² and cosec² identities. The same habits still apply: keep all algebraic roots, filter impossible trig values, and check the original domain.
10 / Your turn
Give all solutions in the stated interval.
02 · Negative cosecant
Solve cosec x = −√2 for −π ≤ x ≤ π.
Hint
Sin x = −√2/2.
Worked solution
x = −3π/4, −π/4. Both lie in the negative half of the interval.
03 · Cotangent zero
Solve cot x = 0 for 0 ≤ x ≤ 2π.
Hint
Set cosine equal to 0 while retaining nonzero sine.
Worked solution
x = π/2, 3π/2. The endpoints 0 and 2π are excluded from the cot domain.
04 · Exact cotangent
Solve cot x = √3 for 0 ≤ x < 2π.
Hint
Tan x = 1/√3.
Worked solution
x = π/6, 7π/6. Cot is positive in Quadrants I and III.
05 · An impossible target
Solve sec x = −1/2 and cosec x = 0 over the real numbers.
Hint
Check the ranges before taking reciprocals.
Worked solution
Neither has a real solution. Sec and cosec have magnitude at least 1 and are never 0.
06 · Endpoints
Solve sec x = 1 for −2π ≤ x ≤ 2π.
Hint
Cos x = 1 at full turns.
Worked solution
x = −2π, 0, 2π. All three are allowed, distinct inputs.
07 · Double angle
Solve cosec(2x) = 2 for 0 ≤ x < π.
Hint
The inner angle covers 0 ≤ 2x < 2π.
Worked solution
2x = π/6 or 5π/6, so x = π/12 or 5π/12.
08 · Shifted cotangent
Solve cot(x − π/4) = 1 for 0 ≤ x < 2π.
Hint
Cot u = 1 at u = π/4 + nπ.
Worked solution
x = π/2 + nπ, giving π/2 and 3π/2. The shifted sine denominator is nonzero at both.
09 · Factor first
Solve cot x(sin x +1) = 0 for 0 ≤ x < 2π.
Hint
Use each factor, then deduplicate.
Worked solution
Cot x = 0 gives π/2 and 3π/2. Sin x = −1 gives 3π/2 again. Complete set: π/2, 3π/2.
10 · Excluded candidate
Solve cosec x = cot x for 0 ≤ x < 2π.
Hint
Both require sin x ≠ 0. Multiply by sine on that domain.
Worked solution
The derived equation 1 = cos x gives x = 0, where sine is 0. Reject it. There are no solutions.
11 · Compare reciprocals
Solve sec x = cosec x for 0 ≤ x < 2π.
Hint
Both sine and cosine must be nonzero.
Worked solution
Clearing denominators gives sin x = cos x, so tan x = 1. Hence x = π/4 or 5π/4; both are allowed.
12 · Squared cosecant
Solve cosec²x = 2 for 0 ≤ x < 2π.
Hint
Cosec x = ±√2, so sin x = ±√2/2.
Worked solution
x = π/4, 3π/4, 5π/4, 7π/4. Keep both signs.
13 · Degrees
Solve cot x = −1 for 0° ≤ x < 360°.
Hint
Tan x = −1; use degree angles.
Worked solution
x = 135° or 315°. The reference angle is 45°.
14 · Reciprocal pair
Solve cosec x + cot x = 2 for 0 ≤ x < 2π, to 4 decimal places.
Hint
Use (cosec x + cot x)(cosec x − cot x) = 1, then add and subtract.
Worked solution
The other combination is 1/2, so cosec x = 5/4 and cot x = 3/4. Thus sin x = 4/5 and cos x = 3/5. The angle must be in Quadrant I: x = arctan(4/3) ≈ 0.9273 rad. Checking the original gives 5/4 +3/4 = 2.
11 / Recap
Solve every branch and check the original equation.
- Sec/cosec equations require a target of magnitude at least 1.
- Cot = 0 is valid and must be handled directly.
- Transform both endpoints for a multiple or shifted angle.
- Factor before dividing by a possibly zero expression.
- Keep both signs when undoing a square.
- Reject excluded candidates and retain allowed endpoints.
- Round numerical angles only after finding the complete set.
Section 1 of 11 · Rewrite the equation and keep the domain