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Choose the complete solution set.

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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

Choose the complete solution set.

cos⁡x=1/2,0≤x<360∘\cos x=1/2,\quad0\le x<360^\circ
  • Degrees; interval half-open.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. {60°}
  2. {60°,300°}
  3. {60°,120°}
  4. {60°,240°}
Working and explanation
BUILD THE REASONING
Hint 1
Use both points with the same horizontal coordinate.
Hint 2
Cosine is positive in quadrants I and IV.
Worked solution
  1. Find the reference angle.
    60∘60^\circ
  2. Reflect across the x-axis.
    360∘−60∘=300∘360^\circ-60^\circ=300^\circ
B: {60°,300°}.
Checks and common pitfalls: The inverse principal value alone omits the reflected solution.
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 ↗