An Interesting Trigonometric Equation

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[xualain]

If both sin(x) and cos(x) are not equal to zero, then
sin(x)^5 < sin(x)^2 …..(1)
cos(x)^7<cos(x)^2 …..(2)
adding (1) and (2)
which is contradictory to what is given that sin(x)^2+cos(x)^7=1
Hence either sin(x)=1 or cos(x)=1 that is
x=2*n*pi+pi/2 or x=2*n*pi

[StaR-uwdc]

nice solution but from the equation sin²x(sin³x-1)=cos²x(1-cos⁵x) we can get the system sin³x-1=0 AND cos²x=0 OR 1-cos⁵x=0 AND sin²x=0. This system has the solution as in video.

[giuseppemalaguti]

Dalle fattorizzazioni si trova sia (1-sinx)=0...(1-cosx)=0...

[RealQinnMalloryu]

{sin^5+sin^ ➖}+{ x+x ➖ }+{cos^7+cos^7 ➖ }+{x+x ➖ sincosx^12^16 sincosx^2^10^2^14 sincosx^1^1^1^1^2^23^2^2 sincosx^1^1^3^1^2 sincosx^3^2 (sincos x ➖ 3sincosx+2).

[jamescollis]

...but are there any other complex solutions? :)

[michaelyap]

Why go so long ? Why not just sin²x(sin³x-1)=cos²x(1-cos⁵x) ?