Quadratic trigonometric equations and complete root sets
導入済み| English | 意味 |
|---|---|
| candidate/ˈkændɪdeɪt/ | 候補解 |
An equation gives two possible sine values. How many angles do they create in one full turn?
- An equation gives two possible sine values. How many angles do they create in one full turn?
- This lesson studies candidate 候选解: A possible solution that still needs checking in the original equation and interval.
Choose the mathematical structure
- 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.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
For 2cos²x-5cosx+2=0, the roots are cosx=1/2 and cosx=2. Why is only 1/2 usable?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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).
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
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)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
sinx=1 gives π/2; sinx=−1/2 gives 7π/6 and 11π/6: three angles.
Test a tempting shortcut
- 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.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Every real root of a quadratic in cosx is an allowed cosine value. This claim is false. Explain which definition or assumption it violates.
How many solutions remain for 2cos²x−5cosx+2=0 on that interval?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Reject cosx=2; cosx=1/2 gives π/3 and 5π/3.
Every real root of a quadratic in cosx is an allowed cosine value.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
Interpret a new situation
- 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.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
How many solutions does sinx cosx=sinx have on that interval?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Factor sinx(cosx−1)=0; x=0 and π, with 0 appearing only once.
Match each trig-equation step to its purpose.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Count ANGLES, not algebraic roots: sin x=1 gives one angle per turn while an interior value like -1/2 gives two.
Use this in your course
- 7357 · A-level · E. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
A possible solution that still needs checking in the original equation and interval. Choose the relationship, show the method, check its assumptions and interpret the result.