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- เผยแพร่เมื่อ 21 ต.ค. 2024
- Math Olympiad | A Nice Trigonometry Equation
Welcome to another exciting Math Olympiad challenge! In this video, we dive into a nice trigonometry equation that will test your understanding and skills. Follow along as we explore various techniques to solve this intriguing problem. Whether you're preparing for a math competition or simply enjoy tackling challenging math problems, this video is perfect for you.
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Topics Covered:
Trigonometry
Trigonometry equations
Trigonometric identities
Math Olympiad
General solution
Math Olympiad Preparation
Math Tutorial
Evaluating trigonometric expressions
Double angle formula
How to solve trigonometric equations
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Thank you for watching!!
It was a wonderful introduction and clearly explained...thanks for sharing Sir 🙏
Since 2x>x we prefer to write 2x First and If sina=sinb then a=2kπ+b or a=2kπ+(π-b). So 2x=2kπ+(x+π/12) ,k is Integer (1) or 2x=2kπ+π-(x+π/12) ,k Integer (2). So 2x-x= 2kπ+π/12 so x=2kπ+π/12 **(3) or 2x+x=2kπ+π-π/12, so 3x=2kπ+π-π/12 so 3x=2kπ+11π/12 so x=2kπ/3+11π/36 **(4), k=0,+-1,+ -2,.... in Both General Solutions (3) and (4). For k=0 (3) is giving x=π/12 and (4) x=11π/36. Other values of k=+-1,+=2 etc. are giving x ooutside of (0,π/2) which are rejected.
(3)^2 ➖ (1)^2/sinx+{3x+3 x➖ }{1x+1x ➖ }/cosox={ 9 ➖ 1}/sinx+{6x^2+2x^2}/cosx={8/sinx+8x^4/cosox} =:16x^4/sinxcosox= 4^4x^4/Sinxcosox 2^22^2x^2^2 1^1^1^1x^1^2; 1^2x (Sinxcosox x ➖ 2sinxcosox x+1).
Πρεπει ημχσυνχ διαφορο του 0. Εχω:(ριζα3-1)συνχ+(ριζα3+1)ημχ=4(ριζα2)ημχσυνχ.ημχ=sinχ και συνχ=cosχ. Εχω ριζα3(συνχ+ημχ)+(ημχ-συνχ)=4(ριζα2)ημχσυνχ=ριζα3[ημ( π/2 -χ)+ημχ]+[ημχ-ημ( π/2-χ)]=2(ριζα2)ημ2χ=...=ριζα3[ριζα2συν( π/4 -χ)]+ριζα2ημ(χ-π/4)=
=ριζα3[ριζα2(συν( π/4)συνχ+ημ( π/4)ημχ)]+ριζα2[ημχσυν( π/4)-συνχ ημ( π/4)]=....=ριζα3(συνχ+ημχ)+(ημχ-συνχ)=2(ριζα2)ημ2χ. Υψωνω στο τετραγωνο τα 2 μελη και εχω 3(1+ημ2χ)+(1-ημ2χ)=8ημ^2(2χ) τελικα 4ημ^2(2χ)-ημ2χ-2=0 θετω ημ2χ=y
4y^2-y-2=0 etc (πολλες πραξεις. Μαλλον κατι δεν παει καλα)