Quadratic trigonometric equations and complete root sets
Введено| English | Значение |
|---|---|
| candidate/ˈkændɪdeɪt/ | кандидат (решение) |
How many angles do they create in one full turn?
- An equation gives two possible sine values.
- This lesson studies candidate 候选解: A possible solution that still needs checking in the original equation and interval.
The method
- Replace one trig function by u, solve the algebraic equation, then solve each valid trig value on the stated interval.
- For sine/cosine, discard u outside [−1,1].
- Tangent accepts every finite real u but excludes angles where cosine is zero.
- Factor before dividing by a trig function, so zero cases are retained.
For 2cos²x-5cosx+2=0, the roots are cosx=1/2 and cosx=2. Why is only 1/2 usable?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The substitution u=cos x carries the range restriction [-1,1]. The algebraic root u=2 has no angle, leaving x=π/3 and 5π/3 from u=1/2 on [0,2π).
A worked case
On 0≤x<2π, solve 2sin²x−sinx−1=0. Put u=sinx: (2u+1)(u−1)=0, so u=−1/2 or 1. The solutions are 7π/6, 11π/6 and π/2; sort them as π/2, 7π/6, 11π/6. For 2cos²x−5cosx+2=0, the algebraic values are 1/2 and 2; discard 2, leaving x=π/3,5π/3. For tan²x−3tanx+2=0, tanx=1 or 2. Thus x=π/4,5π/4,arctan2,π+arctan2, with all four checked in the original equation. For sinx cosx=sinx, factor sinx(cosx−1)=0: x=0 or π. Cancelling sinx would lose π.
Quadratic trigonometric equations and complete root sets
Replace one trig function by u, solve the algebraic equation, then solve each valid trig value on the stated interval
Explain the original interval, denominator or physical reference before using a trig equation.
How many solutions does 2sin²x−sinx−1=0 have on [0,2π)?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
sinx=1 gives π/2; sinx=−1/2 gives 7π/6 and 11π/6: three angles.
Where it goes wrong
- Count angles rather than only algebraic roots: a sine value 1 has one angle per turn, while most interior values have two.
- Reject impossible sine/cosine outputs before taking inverse functions.
- Do not include 2π in a half-open interval, and do not list a repeated endpoint twice.
Not true: Every real root of a quadratic in cosx is an allowed cosine value. Cosine lies between −1 and 1, so a root such as cos x=2 must be rejected.
How many solutions remain for 2cos²x−5cosx+2=0 on that interval?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Reject cosx=2; cosx=1/2 gives π/3 and 5π/3.
Cosine lies between −1 and 1, so a root such as cos x=2 must be rejected.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. Cosine lies between −1 and 1, so a root such as cos x=2 must be rejected.
Applying it
- Use an identity first if both sine and cosine occur: 2sin²x+cosx−2=0 becomes cosx(1−2cosx)=0, giving x=π/2,3π/2,π/3,5π/3 on [0,2π).
- A factorised product is zero when either factor is zero.
- Verify every candidate against the original domain and interval, especially after division or squaring.
How many solutions does sinx cosx=sinx have on that interval?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Factor sinx(cosx−1)=0; x=0 and π, with 0 appearing only once.
Match each trig-equation step to its purpose.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Count ANGLES, not algebraic roots: sin x=1 gives one angle per turn while an interior value like -1/2 gives two.
In your course
- AQA A-level Mathematics (7357), specification section E.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Candidate: A possible solution that still needs checking in the original equation and interval.
- Replace one trig function by u, solve the algebraic equation, then solve each valid trig value on the stated interval.
- Watch for: count angles rather than only algebraic roots: a sine value 1 has one angle per turn, while most interior values have two.