Easiest way to solve this Olympiad problem || Trigonometry || Math Olympiad

In this series, we will regularly update brainstorming problems in mathematics. The collection of problems will be taken from Algebra, Calculus, Mathematical Reasoning, Probability, Number Theory, Functional Analysis, etc.
Subscribe our channel to get regular update. Thank you.

[xiangge]

Or from time 3:00, 1- (sin x)^3 =0, we can get (sin x) = 1 directly since we are discussing real numbers.

[ruslan_]

from 2:16 it can be done in more simple way. from 1st term we've: sinx=0 or sinx=1. from 2nd term we've: cosx=0 or cosx=1. after intersecting we'll have: (sinx=0 and cosx=1) or (sinx=1 and cosx=0). because sinx and cosx can't be = 0 at once. as well as sinx, cosx can't be = 1 at once.

[xiangge]

an easier way can be added and make the proof simpler: Since equation y^2+y+1 = 0 doesn't have a real solution, we can assert that for any real number x, (sin x)^2 + (Sin x) + 1 cannot be zero. The same is correct for (Cos x)

[hilbertonfields]

Why is the expression written after 1:11 true?

[ruslan_]

Let sinx+cosx=t. Then we'lI get an equation: t^5 - 5t+4=0. It has a root t=1 (doubled). After dividing on t-1 two times we've: t^3+2t^2+3t+4=0, which has no solution for t ∈ [-sqrt(2); sqrt(2)].