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

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Topic 3 Trigonometry
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[naakatube]

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

[guti]

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

[KingArkon]

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 😅

[cyruschang]

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

[satrajitghosh]

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)

[maxvangulik]

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

cos(x-45°)=1/sqrt(2)
x-45°={-45°, 45°, 315°}
x={0°, 90°, 360°}

[mxsjncv]

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

[dardoburgos]

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

[Mainak_Goswami]

How about we take sin x = 1-cos x...now use the formula cosx 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

[Mediterranean]

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