Solve the equation sin x +cos x =1 | IB Math

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Опубликовано:

11 сен 2024

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Комментарии : 36

@naakatube Месяц назад

2 sin(x) cos(x) is just sin(2x) !!! Much easier to solve this way!

@guti9709 Месяц назад

Don’t need to watch. Just square both sides!!!

@reload2832 Месяц назад

No, you create extraneous solutions by squaring. Use the harmonic form instead.

@uniquelegend2711 Месяц назад

My sir told square would create false solution Plz can you tell what is that

@rexter8890 Месяц назад

@@uniquelegend2711 squaring trigonometric equations produces extra terms which might give u extra solutions, these extra solutions need not satisfy the original equation, so even after squaring the equation , u must individually check if the solutions are satisfying the equation, if not, then reject the solutions.

@guti9709 Месяц назад

@@uniquelegend2711when u square the left side it’s (sinx + cosx)^2 then u get the 2 terms squared (which equals 1) plus 2sinxcosx (which is equal sin2x) then u solve 2x = 180 and x = 90

@Archimedes_Notes 19 дней назад

@@uniquelegend2711 it will add pi at least which is not a solution

@CalculusIsFun1 19 дней назад

sin(x) + cos(x) = 1 so 1 + sin(2x) = 1 sin(2x) = 0 2x = 0, 2x = 180, 2x = 360, 2x = 540, 2x = 720 x = 0, 90, 180, 270 or 360 180 is invalid and so is 270 so x = 0, x = 180, or x = 360 On 0

@KingArkon Месяц назад

Another solution is to multiply both sides to √2/2 √2/2(sinx + cosx) = √2/2 √2/2sinx + √2/2cosx = √2/2 ✍️ as we know, cos45°=sin45°=√2/2 cos45°sinx + sin45°cosx = √2/2 ✍️ and it looks very similar to: sin(α+β)= sinα*cosβ + sinβ*cosα so we rewrite it as sin(45°+x)=√2/2 ✍️ and as we know, if sinx = a, (a ∈ (0;1) ) the solution is: x = (-1)^k*arcsin(a)+πk, k ∈ Z 45°+x = (-1)^k*45+180°*k, k ∈Z x = (-1)^k*45+180°k - 45°, k ∈Z we are given that x ∈[0;360°] k=0, x =45°-45°= 0✅ k = 1, x = -45+180-45=90°✅ k = 2, x = 45+360-45 = 360°✅ k = -1, x = -45-180-45 = -270 ❌ and no need to check anymore, because none will satisfy the given x I guess this is kinda a bit longer, but just wanted to write ^^ hope, I haven't done any mistakes 😅

@ibmathmaster Месяц назад

I really appreciate your comment and method. But this is a specific curriculum math video (International Baccalaureate SL and HL) where the sum formula sin(α+β)= sinα*cosβ + sinβ*cosα is not studied within the IB SL. you can turn sin x +cos x=1 into sin(x+45)=√2/2 or cos(x-45)=√2/2 using a standardized method in Trigonometry: R*sin(x+-a)or R*cos(x+-a), especially if the question was harder.

@hornkneeeee Месяц назад

@@ibmathmasterWhat's sl and hl?

@ibmathmaster Месяц назад

IB math is an international school curriculum for grades 11 and 12, standard level SL or high level HL

@destruidor3003 Месяц назад

Good explanation at end of the video.

@ibmathmaster Месяц назад

Glad you liked it

@cyruschang1904 Месяц назад

sinx + cosx = 1 (sinx + cosx)^2 = 1^2 = 1 2(sinx)(cosx) = 0 sinx = 0, cosx = 1 or cosx = 0, sinx = 1 x = 2nπ, (4n + 1)π/2

@ChristopherBitti 10 дней назад

You can easily deduce that this can only happen at the right angles (0, pi/2, pi, 3pi/2) through case analysis of the quadrants. In the first quadrant (0, pi/2) the sum is always greater than 1 because the sum of two side lengths of a triangle is always greater than the length of the other side. In the second and fourth quadrants (pi/2, pi) and (3pi/2, 2pi) the sum is always less than 1 because one (either cos(x) or sin(x)) is negative and the other is positive but less than 1. Finally, in the third quadrant the sum is negative. Knowing this, we just need to try these 4 angles. After testing, we can conclude that x = 0, pi/2 and all equivalent angles modulo 2pi.

@Shrutithenerd 20 дней назад

I approached the question like this, 1-cosx = 2sin²x/2 & sinx = 2sinx/2cosx/2, divide both sides by sinx/2 and the equation becomes 2sinx/2 = 2cosx/2, so tan x/2 = nπ+pie/4 & sinx = 0 which is nπ.

@user-mx8sj1nc6v Месяц назад

sinx+tg45*cosx = 1 , multiply both sides by cos45 .... you get sin(x + 45) = cos45 .....

@maxvangulik1988 Месяц назад

cos(x)=(e^ix+e^-ix)/2 sin(x)=-i(e^ix-e^-ix)/2 (1-i)e^ix+(1+i)e^-ix=2 1-i=sqrt(2)e^-ipi/4 1+i=sqrt(2)e^ipi/4 e^i(x-pi/4)+e^-i(x-pi/4)=sqrt(2) cos(x-45°)=1/sqrt(2) x-45°={-45°, 45°, 315°} x={0°, 90°, 360°}

@Mainak_Goswami Месяц назад

How about we take sin x = 1-cos x...now use the formula cosx =1-2sin^x/2.....get 1-cos x = 2sin^2x/2....now we know sinx = 2sinx/2cosx/2...from there you wil get....by combining the equations tan x/2 = 1 or tan pi/4.... we can use the trigonometric equation formula of Tan A = Tan B to get a general sol....now since the period of is 0 to 2pi...I think we can clip the period for tan from (-pi/2,pi/2) that way we take care of the asymptotes of tan x

@Mediterranean81 Месяц назад

sin x + cos x = 1 Sqrt 2*sin (x+Pi/4)=1 Sin(x+Pi/4)=1/sqrt 2 x+Pi/4=Pi/4+2nPi x=0 or 2Pi

@Mediterranean81 Месяц назад

x= Pi/2 (0

@wambertojoseliradequeiroz7919 13 дней назад

+/- kpi/2

@satrajitghosh8162 Месяц назад

sin ( x) + cos ( x) = 1 √ 2 * cos ( x - π /4) =.1 cos ( x - π /4) = 1/√2 = cos (π /4) x - π /4 = 2 n π + π /4, 2 n π - π /4 x = 2 n π + π/2, 2 n π, for imtegral n x = π /2, 0, 2π, 0 are only solution in (0, π)

@Shrutithenerd 20 дней назад

2nπ+π/4 & 2nπ.

@dardoburgos3179 Месяц назад

X= 0, X= π/2. Tiene infinitas soluciones.

@dardoburgos3179 Месяц назад

Para qué elevar al ⬛.

@ManojkantSamal 22 дня назад

By Squaring both sides 1+sin2x=1 Sin2x=0 Sin2x=sin0 2x=0 X=0 degree

@mathboy8188 21 день назад

The 2x = 0 you got is one possibility, but it's not the only possibility. To get all the possibilities, you've should have written: _From sin(2x) = 0 get 2x = 180 n for some integer n._ Continuing on you'd write: Thus x = 90 n for some integer n. And because 0

@MartinNolin-oo9kt 18 дней назад

0:19 360 is not the same as 360 DEGREES. If you only write 360, you don't say anything that it is degrees you mean. You think it is 360 radians, 64800 degrees.