Understand · explore · practise

Quadratic sec, cosec and cot equations

Solve quadratic and fourth-power reciprocal trig equations. Use identities, reject impossible sec and cosec values, retain original domains and find every angle.

Before you startReciprocal trig functions, squared identities and quadratic equations

01 / Two different kinds of root

A quadratic root is a function value, not yet an angle.

Let u stand for one trigonometric function. Solve the algebraic equation for u, check which values that function can take, then find every angle in the requested interval. Each of these steps answers a different question.

Algebraic roots → possible function values → angular solutions

Sec and cosec cannot take values strictly between −1 and 1. Cot can take any real value, including zero. Every function still has excluded angles in its domain.

Values first, angles secondExplore

Let u = sec x. (u − 2)(2u − 1) = 0.

Algebraic roots: 2, 1/2.

Keep 2; reject 1/2 because |sec x| ≥ 1.

x = π/3, 5π/3.

Original domain: cos x ≠ 0.

Green bands show allowed function values. At the filtering stage, green dots are retained and red dots rejected. Gold circle points appear at the final stage. The angles 0 and 2π share a point but are separate endpoints.

EquationStep

Watch an impossible secant value disappear

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Factor a quadratic in secant

Reject a value only for a stated mathematical reason.

Solve 2sec²x − 5sec x + 2 = 0 for 0 ≤ x ≤ 2πWorked example

Let u = sec x: 2u² − 5u + 2 = 0

Factor as (2u − 1)(u − 2).

u = 1/2 or u = 2

These are algebraic roots, not angle answers.

Reject sec x = 1/2

It would require cos x = 2, impossible for a real angle.

For sec x = 2, solve cos x = 1/2

x = π/3 or 5π/3 on the stated interval.

Check both in the original equation

Their cosines are nonzero and 2(2²) − 5(2) + 2 = 0.

A negative secant value is not automatically invalid: values at or below −1 are allowed. It is the magnitude, not positivity, that matters.

03 / Reduce two functions to one

Choose an identity that produces one quadratic variable.

Solve 2cosec²x − 3cot²x = −1 for 0 < x < 2πWorked example

Use cosec²x = 1 + cot²x

Both terms have the same domain sin x ≠ 0.

2 + 2cot²x − 3cot²x = −1

Collect the constant and squared terms.

cot²x = 3, so cot x = ±√3

Cot accepts both real values.

x = π/6, 5π/6, 7π/6, 11π/6

Find all four angles; none has sine zero.

Solve tan²x = 3sec x for 0 ≤ x ≤ 2πWorked example

sec²x − 1 = 3sec x

Use tan²x = sec²x − 1, retaining cos x ≠ 0.

sec x = (3 ± √13)/2

Apply the quadratic formula.

Reject (3 − √13)/2

It lies strictly between −1 and 0.

cos x = (√13 − 3)/2

Take and rationalise the reciprocal of the positive root.

x = α or 2π − α, where α = arccos((√13 − 3)/2)

Use radian mode if finding decimals. Keep this exact form until the end.

04 / Cotangent accepts zero

Do not take a reciprocal before checking whether it exists.

Solve cot²x − cot x = 0 for 0 < x < 2πWorked example

cot x(cot x − 1) = 0

Retain both factors.

cot x = 0 gives cos x = 0 with sin x ≠ 0

x = π/2 or 3π/2.

cot x = 1 gives tan x = 1

Here the reciprocal is legitimate: x = π/4 or 5π/4.

Combine and order the answers

x = π/4, π/2, 5π/4, 3π/2.

Dividing the equation by cot x would lose the zero branch. Replacing cot by 1/tan without checking domains would also hide its values where tangent is undefined.

05 / Reduce a fourth-power equation

Solve for the squared function, then recover both signs.

Solve sec⁴x − 5sec²x + 4 = 0 for 0 ≤ x ≤ 2πWorked example

Let v = sec²x

Then v² − 5v + 4 = (v − 1)(v − 4) = 0.

v = 1 or 4

Both satisfy sec²x ≥ 1.

sec x = ±1 or ±2

Taking square roots creates two signs for each positive v.

cos x = ±1 or ±1/2

Solve all four ordinary cosine equations.

x = 0, π/3, 2π/3, π, 4π/3, 5π/3, 2π

Both endpoints are included; repeated angle answers are listed only once.

If a quadratic in sec²x gives v = −4, reject it before taking roots. If it gives v = 1/4, it is nonnegative but still impossible because sec²x ≥ 1.

06 / Transform the angle interval first

A multiple angle can produce more than two solutions.

Solve 2sec²(2x) − 5sec(2x) + 2 = 0 for 0 ≤ x ≤ 2πWorked example

The same factorisation leaves sec(2x) = 2

The value 1/2 is still impossible.

Let θ = 2x, so 0 ≤ θ ≤ 4π

Do not solve only one turn.

cos θ = 1/2 gives θ = π/3, 5π/3, 7π/3, 11π/3

List every root in the transformed interval.

x = π/6, 5π/6, 7π/6, 11π/6

Divide every angle by 2 and check the original domain cos(2x) ≠ 0.

07 / Retain the original denominator

Cancellation cannot turn an undefined expression into zero.

Solve (sec²x − 4)/(sec x − 2) = 0 for 0 ≤ x ≤ 2πWorked example

Require cos x ≠ 0 and sec x ≠ 2

These are the original restrictions.

Factor the numerator: (sec x − 2)(sec x + 2)

Cancel only on the allowed domain.

The remaining equation is sec x + 2 = 0

The root sec x = 2 from the numerator is excluded.

x = 2π/3 or 4π/3

Both give sec x = −2 and a nonzero denominator.

For a fraction to be zero, its numerator must be zero and its denominator must exist and be nonzero. Multiplying by a denominator is safe only after recording this condition.

08 / Use a reciprocal pair

The product of sec plus tan and sec minus tan is 1.

(sec x + tan x)(sec x − tan x) = 1, if cos x ≠ 0

Given sec x + tan x = 3, find sec x, tan x and the solutions for 0 ≤ x < 2πWorked example

sec x − tan x = 1/3

Divide the identity by the nonzero sum 3.

Add: 2sec x = 3 + 1/3

Sec x = 5/3.

Subtract: 2tan x = 3 − 1/3

Tan x = 4/3.

cos x = 3/5 and sin x = 4/5

Both signs place the angle in Quadrant I.

x = arccos(3/5)

Only this branch has both required signs.

More generally, if sec x + tan x = k, then k ≠ 0, sec x = (k + 1/k)/2 and tan x = (k − 1/k)/2. A negative k is possible; keep its sign rather than taking an unsigned square root.

09 / Count solutions as a parameter changes

The function range and interval endpoints both matter.

How many solutions does (sec x − k)(2sec x − 1) = 0 have for 0 ≤ x < 2π?Worked example

The second factor requires sec x = 1/2

This branch never supplies a real angle.

For |k| < 1 there are no solutions

These values are outside the secant range.

For k = 1 or k = −1 there is one solution

They are x = 0 and x = π respectively; 2π is excluded.

Can sec x + tan x = 0 hold for a real angle where both functions exist?

Hint

Multiply by sec x − tan x.

Worked solution

No. The product would be 0, but the identity says it must be 1. This also explains the restriction k ≠ 0 in the reciprocal-pair formulas.

11 / Recap

An algebraic solution must survive every later check.

Use identities to reduce to one trig function.
Solve the quadratic, or substitute a squared function for a fourth-power equation.
Sec and cosec require magnitude at least 1; cot accepts every real value.
Recover both signs after solving for a squared function.
Find every angle in the correct transformed interval.
Keep original denominator exclusions and check endpoints.

Section 1 of 11 · Two different kinds of root