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When 0≤x<360°, what interval must be searched for 2x?

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高一必修 第一册(A版).pdf · 5.4 · PDF 203 / printed page 196

Revisit first: Unit-circle coordinates and signsIdentity choice and domain preservationAmplitude, period and phase in models

TOPIC 01

Complete trigonometric solution sets

Solve a transformed angle, recover every branch and intersect the original interval.

What you will be able to explain

  • Retain all angle branches and periodic copies in the requested interval.
  • Detect lost solutions from division and extraneous solutions from squaring.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Your turn

When 0≤x<360°, what interval must be searched for 2x?

u=2xu=2x
  • Degrees.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. [0,180°)
  2. [0,360°)
  3. [0,720°)
  4. [0,720°]
Working and explanation
BUILD THE REASONING
Hint 1
Multiply both endpoint bounds by positive 2.
Hint 2
Keep the endpoint openness.
Worked solution
  1. Transform the lower and upper bounds.
    0≤2x<720∘0\le2x<720^\circ
  2. The upper endpoint remains excluded.
    u∈[0,720∘)u\in[0,720^\circ)
C: [0,720°).
Checks and common pitfalls: Keeping [0,360°) would miss one phase cycle.
Think first. Reveal a hint when the class is ready.
Focus on one question
Teacher preparation and assessment
40-minute lesson
MinutesFocusTeaching and student action
5Branch diagnosisUse diagnostics to distinguish principal values and transformed intervals.
8Build solution setsMark equal-height points and periodic copies before filtering.
10Audit operationsPairs contrast safe division in sin x=cos x with lost roots in sin 2x=cos x.
12Classify and modelGroups classify parameter root counts and translate the wheel threshold.
5Exit completenessRequire a complete set and a reason no other branch fits.
Question sequence
  • What interval does the transformed angle actually traverse?
  • Did this operation lose or add candidates?
Board development
  • Preserve factor branches.
    cos⁡x(2sin⁡x−1)=0\cos x(2\sin x-1)=0
  • Map an angle interval back to time.
    π/6<πt/4<5π/6\pi/6<\pi t/4<5\pi/6
Expected correct and incorrect responses
  • “Both endpoints count because they are different real x-values” correctly handles k=0.
  • “Divide by cosine immediately” risks losing valid zero-factor solutions.
Follow-up questions
  • Why does substituting all found roots not prove that none were missed?
Remediation
  • For missing branches, draw the horizontal value line before invoking inverse trigonometry.
  • For transformed-interval errors, put x bounds and angle bounds in separate columns.
Exit check
  • Solve cos 2x=1 for 0≤x≤2π. Expected 0,π,2π with closed endpoints.
Assessment checklist and sample responses
  • 2 marks: branch and domain completeness.
    Full: four roots for sin 2x=1/2 on [0°,360°); partial: first two only.
  • 2 marks: validate nonreversible steps.
    Full: reject positive-sine roots after squaring; partial: accept all four squared-equation roots.
No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.
Curriculum and source notes ↗