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Quadratic trigonometric equations

Solve quadratic equations in sine, cosine and tangent. Use identities, keep negative and zero roots, reject impossible values and find all radian angles.

Before you startQuadratic factorisation, trigonometric identities and complete interval solutions

01 / Solve the algebra before finding angles

A trigonometric value is an intermediate answer.

In 2sin²x−sin x−1=0, treat sin x as one quantity u. Solve 2u²−u−1=0 first, then solve the separate trigonometric equations. Each allowed u may give more than one angle.

Rewrite in one trig function → solve the quadratic → check allowed values → find all angles → check the original domain.

Change the substitution in the model. A value rejected by sine or cosine can still be a valid tangent value.

Algebraic roots, then anglesExplore

(2u + 1)(u − 1) = 0.

Algebraic roots: −1/2 and 1; both allowed.

3 solutions: π/2, 7π/6, 11π/6.

Green points are allowed trigonometric values; red points are rejected. A repeated algebraic root is used once. The final interval is 0≤x<2π.

SubstitutionQuadratic equation

Watch algebraic roots become angle solutions

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

02 / Factor a quadratic in sine

Keep both factors and both trigonometric branches.

Solve 2sin²x−sin x−1=0 for 0≤x<2πWorked example

Let u=sin x: 2u²−u−1=(2u+1)(u−1)

The algebraic roots are u=−1/2 and u=1.

Both lie in [−1,1]

A negative sine value is allowed.

sin x=−1/2 gives 7π/6, 11π/6

Use quadrants III and IV.

sin x=1 gives π/2

This extreme value gives one root per turn.

x=π/2, 7π/6, 11π/6

Three distinct angle solutions.

03 / Reject impossible values, not negative ones

Sine and cosine lie in [−1,1]; tangent can take any real value.

Solve cos²x−cos x−2=0 on 0≤x<2πWorked example

(cos x−2)(cos x+1)=0

The algebraic candidates are 2 and −1.

Reject cos x=2

It is outside cosine’s range.

Keep cos x=−1

A negative value is not itself a reason to reject it.

x=π

Only one angle in this interval.

By contrast, tan²x−4=0 has tan x=±2. Both values are allowed. If α=arctan 2, its one-turn roots are α, π−α, π+α and 2π−α. Tangent’s vertical asymptotes remain excluded inputs.

04 / Use an identity to obtain one function

Substitution works after the equation has one kind of trig value.

Solve 3cos²x+2sin x−3=0 on 0≤x<2πWorked example

cos²x=1−sin²x

This identity holds for every real x.

3(1−sin²x)+2sin x−3=0

Expand before factorising.

sin x(2−3sin x)=0

The values are sin x=0 or sin x=2/3.

x=0, π, α, π−α, where α=arcsin(2/3)

For an ordered list use 0, α, π−α, π.

Choose the identity that matches the unsquared term. An equation containing cos²x and sin x is usually easiest in sine; one containing sin²x and cos x is usually easiest in cosine.

05 / A squared trig value has two signs

Taking only the positive square root loses solutions.

Solve sin²x=1/4 on 0≤x<2πWorked example

sin x=±1/2

Square roots introduce both signs.

Positive branch: π/6, 5π/6

Quadrants I and II.

Negative branch: 7π/6, 11π/6

Quadrants III and IV.

Four distinct solutions

Substitution into the squared equation confirms every one.

Similarly, tan²x=3 gives tan x=±√3 and x=π/3, 2π/3, 4π/3, 5π/3 on one turn.

06 / Factor without dividing away a solution

Dividing by a trig function assumes it is nonzero.

Solve 2sin²x=sin x on 0≤x<2πWorked example

2sin²x−sin x=0

Bring every term to one side.

sin x(2sin x−1)=0

Use the zero-product rule.

sin x=0 or sin x=1/2

The zero branch must remain.

x=0, π/6, 5π/6, π

Dividing by sin x at the start would lose 0 and π.

07 / Repeated roots and no real roots

Algebraic multiplicity is different from the number of angles.

Solve (2cos x−1)²=0 on 0≤x<2πWorked example

2cos x−1=0

The repeated factor produces the single trig value 1/2.

x=π/3 or 5π/3

List each angle once.

If the substituted quadratic is u²+1=0, it has no real u, so there is no real angle solution. If it is u²−4=0, real u exist but neither is a sine or cosine value. These are different reasons for an empty solution set.

08 / A quadratic can contain a multiple angle

Filter values first, then use the full inner-angle interval.

Solve 2sin²(2x)−sin(2x)−1=0 for 0≤x<πWorked example

Let u=sin(2x)

u=1 or u=−1/2.

Set θ=2x, so 0≤θ<2π

θ is an angle; u is a function value. Do not confuse them.

θ=π/2, 7π/6, 11π/6

Solve the two allowed sine equations.

x=π/4, 7π/12, 11π/12

Divide all inner-angle roots by 2.

09 / Keep the original denominator restrictions

Cancellation changes the expression’s appearance, not its domain.

Solve (2sin²x−sin x)/(sin x)=1 for 0≤x<2π. The denominator requires sin x≠0 before any cancellation.

Cancel only on the stated domainWorked example

2sin x−1=1, with sin x≠0

Cancellation is valid at the allowed inputs.

sin x=1

This satisfies the nonzero requirement.

Solve cos²x=1 on −π≤x≤π.

Hint

cos x=±1; include both endpoints.

Worked solution

x=−π,0,π.

11 / Recap

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

Use identities to express the equation in one trig function.
Factor and retain both algebraic branches.
Keep valid negative and zero values.
Reject values outside sine/cosine’s range and nonreal values.
Find all distinct angles over the correct interval.
Preserve original denominator restrictions.

Section 1 of 11 · Solve the algebra before finding angles