Solving A Trigonometric Equation

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[@heet888]

You always come up with these interesting problems

[@SyberMath]

Glad to hear that!

[@matthewfeig5624]

If I see cos x +/- sin x, I think to shift the variable by pi/4 so you can write them as a single trig function. Just multiply by sqrt(2)/2 and think about the sum identities for sine and cosine. LHS = cos x + sin x = sqrt(2)*sin(x+pi/4) And cos(2x) can also be expressed nicely in terms of x+pi/4. RHS = cos(2x) = sin(2x+pi/2) = sin(2(x+pi/4)) So in terms of y=x+pi/4, the equation is sqrt(2) sin y = sin(2y) = 2*sin y*cos y That can be factored to give sin y = 0 and cos y = sqrt(2)/2.

[@honestadministrator]

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

[@NadiehFan]

I thought of a third method, but to see this, you will have to change the sorting order of the comments to _newest first._

[@Katzenkratscher]

Interesting. Unit circle has corners. I have never thought about that.

[@SyberMath]

Unit square! 😜😜

[@Nobodyman181]

U solve very interesting problems❤

[@SyberMath]

Thank you!

[@jaimeduncan6167]

Excellent. Why cos(x)-sin(x)=1 has so few solutions needs justification. Clearly, sin(x)=-0.5 and cos(x)=0.5 seems to be a solution. The reason that will not work is the Pythagorean identity but it needs justification. That is the advantage of the first method.

[@scottleung9587]

I used the second method.

[@mikeschieffer7899]

sinx + cosx = cos(2x) sinx + cosx = (cosx)^2 - (sinx)^2 (sinx)^2 + sinx = (cosx)^2 - cosx Completing the square on both sides gives: (sinx)^2 + sinx + 1/4 = (cosx)^2 - cosx + 1/4 (sinx +1/2)^2 = (cosx - 1/2)^2 Taking square root of both sides gives two equations: Eq1: sinx + 1/2 = cosx - 1/2 and Eq2: sinx + 1/2 = -(cosx - 1/2) Eq1: sinx - cosx = -1 sqrt(2)*cos(x - 3*pi/4) = -1 cos(x - 3*pi/4) = -1/sqrt(2) = -sqrt(2)/2 Taking arccos of both sides gives: x - 3*pi/4 = +/- 3*pi/4 + 2*pi*n x = 0 or 3*pi/2 + 2*pi*n Eq2: sinx + cosx = 0 sqrt(2)*cos(x - pi/4) = 0 cos(x - pi/4) = 0 Taking arccos of both sides gives: x - pi/4 = +/- pi/2 + 2*pi*n x = 3*pi/4 or -pi/4 + 2*pi*n

[@liahsheep]

0:15 where's the possibility of the 3rd method?

[@SyberMath]

Check the highlighted comment by @NadiehFan

[@Dae-Ying-Kim12345]

* what about sin(x)+coc(x)=tan(x) ? *

[@mk-ot6wp]

Solve this: Sin4x +tanx =0

[@DonEnsley-yi2ql]

problem sin x + cos x = cos 2x sin x + cos x = cos² x - sin² x sin x + cos x = (sin x + cos x)(cos x - sin x) (sin x + cos x)[1-(cos x - sin x)] = 0 By zero product property, solutions from sin x + cos x = 0 sin x = - cos x √(1-cos² x) = - cos x 1- cos² x = cos² x 2 cos² x = 1 cos x = ± 1/√2 x = π/4, 3π/4, 5π/4, 7π/4 sin x + cos x = 0 But this is true only when sin x = - cos x This is true for x = 3π/4, 7π/4 x = 3π/4, 7π/4 = 3π/4+nπ, n integer solutions from cos x - sin x = 1 cos x - 1 = √(1-cos² x) cos² x-2 cos x + 1 = 1 - cos² x 2 cos² x-2 cos x = 0 cos x (cos x - 1) = 0 cos x = 0 x = πn/2 cos x = 1 x = 2πn All solutions: x ∈ { ¾ π+nπ, 2πk, ½ π m, (n,k,m ∈ ℤ) }

[@VlogByMijan]

Please make videos on hardest IMO algebra problems 🙏🙏🙏🙏🙏. This will be more helpful and it will also increase views of your channel.

[@heet888]

They are kind not in domain of the kind of solution this guy does.(BTW got gold in IMO 2023 Class IX)