# Solution 82: Simple Bashing & Cauchy-Schwarz "Equality"

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**Published on Mar 7, 2019**- We efficiently solve the trigonometric problem in two different ways, with a subtle use of Cauchy-Schwarz inequality for the second method.

Congratulations to Rishav Gupta, JIN ZHI PHOONG, Hiren Bavaskar, Serengeti Ghasa, mstmar, adandap, EyyLmaoo, UbuntuLinux, Intergalactic Bajrang dal, and Bintang Alam Semesta W.A.M for successfully solving this math challenge question! Rishav Gupta was the first person to solve the question.

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Super bike And sport2 months agoI love ❤️

K Design3 months agoHello

can you solve this ? find gamma or beta ; we have only x , y, and alpha

images2.imgbox.com/87/a4/LQ92SuQW_o.png

Aaron He3 months ago^{+2}Hi, what is your experience with the art of problem solving volume one book? I am considering to buy it in order to improve my problem solving skills.

LetsSolveMathProblems3 months ago^{+2}AoPS Volume 1, quite simply, is outstanding. It would be one of the first books I would recommend to anyone wishing to build or solidify one's foundation for mathematical problem solving. I personally learned much from Volume 1, as well. =)

If you have already mastered most of the material in Volume 1, I would strongly recommend checking out Volume 2 (of course!), Paul Zeitz's "Art and Craft of Problem Solving," and Arthur Engel's "Problem Solving Strategies." All three books are superb, although I do warn you that Engel's book requires a certain degree of mathematical maturity.

Atharva #breakthrough3 months agohow do u solev it to find this x? value?

traso3 months ago^{+3}I don't take math classes anymore so your videos keep me fresh

David Franco3 months ago4:50 :o oh that is the part were I got stuck, I did not recognize the perfect square -.-"

Shanmuga Sundaram3 months agoDid you prove the uniqueness of "x" in the said interval?

Marco Dalla Gasperina3 months ago^{+1}sinx is monotonic (positive) on [0,pi/2] and so is sin^4(x). cos(x) is monotonic (negative) and so is cos^3(x). Therefore, if they intersect, they can only intersect at one point.

Davide3 months agoWhere would you even need the uniqueness of x?

adandap3 months agoNot a proof, but Mathematica finds a single root in the interval at ~ 0.835. (In fact it's also a local minimum.) It can also find a closed expression for it, but it's not pretty!

x = 2 ArcCos[\[Sqrt](5/8 + 1/( 8 Sqrt[3/(19 + (4960 - 96 Sqrt[849])^(1/3) + 2 2^(2/3) (155 + 3 Sqrt[849])^(1/3))]) -

1/2 \[Sqrt](19/24 - 1/48 (4960 - 96 Sqrt[849])^(1/3) - 1/12 (1/2 (155 + 3 Sqrt[849]))^(1/3) + 9/8 Sqrt[3/(19 + (4960 - 96 Sqrt[849])^(1/3) +

2 2^(2/3) (155 + 3 Sqrt[849])^(1/3))]))]

Pranjal Das3 months ago^{+1}So thanks for the video

Pranjal Das3 months ago^{+1}I didn't know the Cauchy's inequality

Davide3 months agoPranjal Das bj hahaha

Pranjal Das3 months ago^{+1}That's cool

adandap3 months ago^{+4}I do like me a bit of straightforward bashing! That's my default setting. :)

el tapa3 months agoToo Easy for me :p

Great edition

HARRIS LAM3 months agoBeautiful