01 · Two sine branches
2sin² x − sin x − 1 = 0
Hint
(2s + 1)(s − 1) = 0.
Worked solution
sin x = −1/2 or 1
x = 90°, 210°, 330°
Understand · explore · practise
Quadratic trigonometric equations
Solve quadratics in sin, cos and tan, use identities, keep zero factors and reject impossible or extraneous solutions. Includes transformed angles and biquadratic extensions.
Before you startQuadratic equations, identities and complete trigonometric solution families
01 / A quadratic in a ratio
Use algebra first, then solve each trig branch.
In 2sin² x + sin x − 1 = 0, substitute s = sin x. This gives 2s² + s − 1 = 0, which factorises as (2s − 1)(s + 1) = 0.
sin x = 1/2 or sin x = −1
For 0° ≤ x < 360°:
x = 30°, 150°, 270°
Each permitted algebraic root starts a separate trig equation. A double algebraic root does not mean an angle should be listed twice. The model keeps each branch and its angle solutions visible.
Keep every valid branchChoose and compare
2sin² x + sin x − 1 = 0 on [0°,360°). (2sin x − 1)(sin x + 1) = 0. sin x = 0.5: 30°, 150°. sin x = -1: 270°. Complete sorted list: 30°, 150°, 270°. Non-exact angles are rounded to four decimal places.
Watch a quadratic split into two trigonometric branches
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Reject impossible ratios
A real algebraic root may be an impossible cosine.
Solve 2cos² x − 3cos x − 2 = 0 on 0° ≤ x < 360°. Factorising gives:
(2cos x + 1)(cos x − 2) = 0
cos x = −1/2 or cos x = 2
Reject cos x = 2 because cosine lies in [−1,1]. The remaining branch gives x = 120°, 240°. Tangent has no comparable [−1,1] restriction: tan x = 2 is valid.
No real algebraic roots
For 2sin² x + sin x + 2 = 0, the quadratic discriminant is 1 − 16 = −15. There is no real value of sin x and hence no real angle solution. Completing the square reaches the same conclusion.
03 / Use an identity first
Express the equation in one ratio.
Solve 3cos² x + 2sin x − 3 = 0 for 0° ≤ x ≤ 360°. Replace cos² x by 1 − sin² x.
3(1 − sin² x) + 2sin x − 3 = 0
sin x(2 − 3sin x) = 0
sin x = 0 or sin x = 2/3
The first branch gives 0°, 180°, 360°. The second gives α = sin⁻¹(2/3) and 180° − α. To one decimal place, the complete sorted answer is:
x = 0°, 41.8°, 138.2°, 180°, 360°
The identity must use the same full argument. It does not let you replace cos² 2x with 1 − sin² x.
04 / Keep zero factors
Dividing by sine could delete solutions.
Solve 3sin x = tan x for 0° ≤ x ≤ 360°. The original tangent requires cos x ≠ 0. Replace tan x by sin x/cos x and multiply by that non-zero denominator.
3sin x cos x = sin x
sin x(3cos x − 1) = 0
Keep both branches: sin x = 0 gives 0°, 180°, 360°; cos x = 1/3 gives approximately 70.5°, 289.5°. All satisfy the original domain. Dividing by sin x at the start would lose the three zero-sine answers.
Factor by grouping
3sin x cos x − cos² x + 3sin x − cos x
= (cos x + 1)(3sin x − cos x)
Set each factor to zero. On 0° ≤ x < 360°, cos x = −1 gives 180°. In the other branch, cos x = 0 cannot work, so tan x = 1/3 gives approximately 18.4°, 198.4°.
05 / Tangent quadratics
Keep both positive and negative algebraic roots.
For tan² x − tan x − 2 = 0:
(tan x − 2)(tan x + 1) = 0
tan x = 2 or tan x = −1
On 0° ≤ x < 360°, the first branch gives approximately 63.4°, 243.4°. The second gives 135°, 315°. Sort the four answers before finishing.
Similarly, tan² x = 3 means tan x = ±√3, not just √3. On a full half-open turn this gives 60°, 120°, 240°, 300°.
06 / Squared shifted angles
Solve the algebra, then map the argument interval.
Solve sin²(2x + 30°) = 1/4 for 0° ≤ x < 180°. Let u = 2x + 30°, giving 30° ≤ u < 390°.
sin u = ±1/2
u = 30°, 150°, 210°, 330°
x = (u − 30°)/2 = 0°, 60°, 90°, 150°
The candidate u = 390° is excluded by the open upper endpoint. Keep the ± step separate from the two angle families for each sign.
07 / Original denominators
Cross-multiplication needs an exclusion list.
Solve (2cos x + 1)/(1 − cos x) = 1 for 0° ≤ x ≤ 360°. Before multiplying, exclude cos x = 1, so x = 0°, 360° cannot be solutions.
2cos x + 1 = 1 − cos x
3cos x = 0
x = 90°, 270°
Both surviving angles have denominator 1 and satisfy the original equation. Do not exclude them just because tangent would be undefined: this expression does not contain tangent.
Cancellation can leave a hole
For sin² x/(1 − cos x) = 2, require cos x ≠ 1. The identity sin² x = (1 − cos x)(1 + cos x) reduces the equation to 1 + cos x = 2. This demands cos x = 1, precisely the excluded case. There are no solutions.
08 / Check after squaring
A squared equation forgets which sign was required.
Solve tan x = cos x on 0° ≤ x < 360°. Its domain requires cos x ≠ 0. Multiplying by cosine gives sin x = cos² x, so sine must be non-negative.
sin x = 1 − sin² x
s² + s − 1 = 0
s = (−1 ± √5)/2
The negative algebraic root is below −1 and is impossible. Write r = (√5 − 1)/2. The solutions are x = sin⁻¹ r and 180° − sin⁻¹ r, approximately 38.2°, 141.8°.
If instead you square sin x = cos² x, angles with sin x = −cos² x also satisfy the new equation. They fail the original sign requirement. Substitution in the original equation removes these extraneous candidates.
09 / Extension: fourth powers
Use a second substitution and take both square roots.
Solve tan⁴ x − 5tan² x + 4 = 0 on 0° ≤ x < 360°. Let v = tan² x ≥ 0.
v² − 5v + 4 = (v − 1)(v − 4) = 0
tan² x = 1 or 4
tan x = ±1 or ±2
There are eight distinct answers. With α = tan⁻¹ 2 ≈ 63.4349°:
x = 45°, α, 180° − α, 135°,
225°, 180° + α, 360° − α, 315°
Written to one decimal place in increasing order: 45°, 63.4°, 116.6°, 135°, 225°, 243.4°, 296.6°, 315°. Reject any negative value of v before taking a square root.
10 / Your turn
Make a branch list and check it against the original equation.
Use 0° ≤ x < 360° unless an interval is stated otherwise. Round non-exact answers to one decimal place.
02 · Reject a root
2cos² x + 3cos x − 2 = 0
Hint
(2c − 1)(c + 2) = 0.
Worked solution
cos x = 1/2 or −2
Reject −2; x = 60°, 300°
03 · Mixed squares
2cos² x + sin x − 1 = 0
Hint
Replace cos² x with 1 − sin² x.
Worked solution
2sin² x − sin x − 1 = 0
sin x = 1 or −1/2
x = 90°, 210°, 330°
04 · Preserve zero
2sin x cos x = sin x
Hint
Factor sin x rather than divide by it.
Worked solution
sin x(2cos x − 1) = 0
x = 0°, 60°, 180°, 300°
05 · Tangent squared
tan² x = 1/3
Hint
tan x = ±√3/3.
Worked solution
x = 30°, 150°, 210°, 330°
06 · A shifted square
cos²(x − 30°) = 1/4
Hint
−30° ≤ u < 330° and cos u = ±1/2.
Worked solution
u = 60°, 120°, 240°, 300°
x = 90°, 150°, 270°, 330°
07 · A forbidden answer
sin² x/(1 − cos x) = 2
Hint
Start by excluding cos x = 1.
Worked solution
The simplified equation requires cos x = 1, which makes the original denominator zero. No solutions.
08 · Repeated root
4sin² x − 4sin x + 1 = 0
Hint
(2sin x − 1)² = 0.
Worked solution
sin x = 1/2
x = 30°, 150°
List each angle once.
09 · Squaring adds candidates
A student squares sin x = cos² x and accepts x ≈ 218.2°. Why does it fail?
Hint
Compare the signs on the two sides of the original equation.
Worked solution
At that angle sine is negative but cosine squared is non-negative. The squared equation lost the original sign requirement, so this candidate must be rejected.
10 · A closed negative interval
sin² x = 1, −450° ≤ x ≤ 90°
Hint
sin x can be 1 or −1.
Worked solution
x = −450°, −270°, −90°, 90°
Both closed endpoints satisfy the equation.
11 / Recap
Algebra produces branches; the original equation decides.
- Substitute a single ratio and solve the algebraic equation.
- Sine and cosine must lie in [−1,1].
- A squared ratio needs both signs unless a restriction removes one.
- Factor instead of dividing by an expression that may be zero.
- Carry original denominator exclusions through the working.
- Check squared-equation candidates in the original equation.
- Map any multiple or shifted argument over its full interval.
For a foundation refresher, revisit solving quadratics or trigonometric identities.
Section 1 of 11 · A quadratic in a ratio