Viral Math Olympiad Questions | How to solve Algebra - Nice Math Olympiad Problem
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- เผยแพร่เมื่อ 29 มิ.ย. 2024
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🎉🎉🎉🎉🎉🎉🎉
Very Very useful video 🎉🎉🎉🎉🎉
Thanks a lot
Fantastic 🎉🎉🎉🎉🎉
Thanks 🤗
Very very informative video 🎉🎉🎉🎉🎉
Glad you think so!
Glad you think so!
*=read as square root
^=read as to the power
As per question
4^(sinx)^2+4^(cos x)^2=3.*2
Let us explain
4^(cosx)^2=4^(1-sinx^2)
=(4^1)/(4^sinx^2)=4/(4^sinx^2)
Suppose 4^sinx^2=Y
So the given question can be written as below
Y+(4/y)=3.*2
(Y^2+4)/y=3.*2
Y^2+4=3.*2.y
Y^2- 3.*2y+4=0
Here a=1, b=-3.*2, c=4
Putting the formula (-b± *(b^2-4ac)/2a, we shall get
Y=-(-3.*2)±*{(3.*2)^2-4.1.4)}/2.1
Y=3.*2±*{(18-16)}/2
Y={3.*2±*2}/2
Y=*2{(3±1)}/2
Y=(*2.4)/2 and( *2.2)/2
Y=2.*2 and *2
Y=(*2)^3 and (*2)^1
According to the 'suppose'
Y=4^(sinx^2), {4=(*2)^4}
Y=*2^(4sinx^2)
Now put the value of Y
*2(4sinx^2)=(*2)^3
4 sinx^2=3
Sinx^2=3/4
Sinx=±(*3/2)
Sinx =sin60
X=60 degree
Again
(*2)^4sinx^2=(*2)^1
4sinx^2=1
Sinx^2=1/4
Sinx=±1/2
Sinx =sin30
X=30 degree
(N:B: There may be another pair of answers for negative values )
Nice
Couldn't figure out to use the degree indicator with 30, but did for 60. Oy.
Commas in description are wrong.Remove comma immediately okay 🎉🎉🎉🎉🎉
According to youtube policy hashtag should not be more than 15 🎉🎉🎉🎉🎉
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