@@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!
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. 😂🤣
-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
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.
(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.
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
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".
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@3:00, I think you confused "sine" with "cosine". sine of 90 is 1 whereas cosine of 90 is 0.
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).
Copy paste not allowed !
Equation gives : (tan(x))^3 = -1 ----> tan(x) = -1 ----> x = -π/4 + n*π
Why do you freak about copy and paste? Ok, Fine. You got the greatest comment
@@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!
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. 😂🤣
Nice job!
Thanks!
-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
5:13 thank you for this info .🤩
😊
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.
(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
(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.
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.
tan^3 x = -1
tan x = -1
x = nπ - π/4
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
short and sweet!
👍
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".
Cubic? More like "Cool and lit!" 🔥
Thank you!
x=-π/4+πn, n→Z
Answer: 2.3562, 3.927
You must be joking! Solving tan^3 x=-1 for almost 10 mins!