A Problem From ARML-NYSML Math Contests

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[@FisicTrapella]

cos x = - sin x
If sin x = cos x, that gives us x = (pi)/4.
The opposite sign means we are in the 2nd or 4th quadrant
Then, x = 3(pi)/4 + n(pi).

[@lawrencejelsma8118]

Finally my method that is so trivial 😂 Without the assumption cos(x) = 0, Divide both sides of the equation by cos^3(x) and multiply both sides of rhe equation to get tan^3(x) = -1. The second and third quadrants have negative tangent values. Evaluate parallelly as in the video from the tangent of an angle = -1 from the quadrant 2 at 3π/4 primary angle analysis in tangent graphs. From Tan(x) or COT(x) function graphs mathematics avoid like a plague to prefer Sin(x) and Cos(x) continuous function properties in all x values. 😂🤣

[@vighnesh153]

@3:00, I think you confused "sine" with "cosine". sine of 90 is 1 whereas cosine of 90 is 0.

[@WahranRai]

Copy paste not allowed !
Equation gives : (tan(x))^3 = -1 ----> tan(x) = -1 ----> x = -π/4 + n*π

[@CriticSimon]

Why do you freak about copy and paste? Ok, Fine. You got the greatest comment

[@WahranRai]

@@CriticSimon Some wait for others' responses to submit their responses by copying others' work.
If you want to send a response send your own creation and don't plagiarize that of others!

[@joyneelrocks]

For the cos(x) + sin(x) = 0 case, just square both sides and life is simple:
cos^2(x) + 2cos(x)sin(x) + sin^2(x) = 0
sin(2x) + 1 = 0
2x = -π/2 + 2πn, where n in Z
So x = -π/4 + πn

[@ianfowler9340]

short and sweet!

[@davidbass5115]

-sin^3(b)=sin^3(-b)
sina=cos(90-a)
plug in
cos^3x = cos^3(90+x)
take cube root
x=90+x+360k -> no sol I think
or
x=-(90+x)+360k
2x=-90+360k
x=-45+180k
in radians
x=-pi/4+pik
comes out to

[@seanfraser3125]

(cosx)^3 + (sinx)^3 = 0
-> (cosx + sinx)((cosx)^2 + cosxsinx + (sinx)^2) = 0
-> (cosx + sinx)(1 + cosxsinx) = 0
-> cosx + sinx = 0 or 1+ cosxsinx = 0
If 1 + cosxsinx = 0, then cosxsinx = -1. Thus sin2x = -2, which is impossible. So this gives no solutions.
If cosx + sinx = 0, then cosx = -sinx. Since sin and cos are both 0 for the same x, x must be a radian value in the second or fourth quadrant. For the second quadrant, we have 3pi/4 + 2npi. For the fourth quadrant, we have -pi/4 + 2npi. Combining these, we have the set of solutions x = -pi/4 + npi.
So x = -pi/4 + npi for any integer n.

[@scottleung9587]

Nice job!

[@SyberMath]

Thanks!

[@ianfowler9340]

Whoops. sin(pi/2) = 1 not 0.
Any linear combination of sine and cosine with the same period can be written as a single sine or cosine.
sin(x) + cos(x) = 0
sqrt(2) sin( x + pi/4) = 0
x +pi/4 = k*pi
x = - pi/4 + k*pi or if you prefer x = 7*pi/4 + k*pi as some prefer to have k=0 produce an angle between 0 and 2pi.

[@barakathaider6333]

👍

[@honestadministrator]

(cos ( x) + sin ( x))
* ( 1 - cos ( x) sin ( x)) = 0
cos ( x - π /4) = 0, sin (2x) = 2
x = π /4 + ( n + 1/2) π = ( n + 3/4) π

[@rob876]

tan^3 x = -1
tan x = -1
x = nπ - π/4

[@PunmasterSTP]

Cubic? More like "Cool and lit!" 🔥

[@SyberMath]

Thank you!

[@AmirgabYT2185]

x=-π/4+πn, n→Z

[@robot8324]

5:13 thank you for this info .🤩

[@SyberMath]

😊

[@rakenzarnsworld2]

Answer: 2.3562, 3.927

[@bjorntorlarsson]

I had a course mate in econometrics (maths for economists, so bah!) And we students in a group had great difficulties solving a maths problem. We were all very surprised when the stupidest of us claimed to have found the solution! Until we by a glance saw that he assumed 2=0. Apropo the "we're solving for X here".

[@tixanthrope]

Am I mssing sthg here?
the original equation directly implies that
cos x = -sin x
which implies x = 3 pi / 4 + n pi, where n is any integer.

[@vladimirkaplun5774]

You must be joking! Solving tan^3 x=-1 for almost 10 mins!

[@DonEnsley]

problem
cos³x = - sin³x
cos³x + sin³x = 0
A sum of two cubes this is.
Factor.
(cos x + sin x)(cos²x - cos x sin x + sin²x) = 0
(cos x + sin x)(1 - cos x sin x) = 0
By the zero propuct property,
cos x = - sin x
tan x = -1
x = 3π/4 + πk
or
1 = cos x sin x
sec²x = tan x
tan²x - tan x + 1 = 0
x = arctan(½ ± i ½√3)
answer
x ϵ {3π/4+πk, (n ϵ ℤ)
arctan(½-i ½√3),
arctan(½+i ½√3) }