trig q. (1 Viewer)

[jake2.0]

solve

sinx + cosx = 1 + sin2x

x is between 0 and 2pi (inclusive)

thanks guy

 

[香港!]

solve

sinx + cosx = 1 + sin2x

x is between 0 and 2pi (inclusive)

thanks guy

sinx + cosx = 1 + sin2x
sinx+cosx=1+2sinxcosx
sinx+cosx-2sinxcosx=1
sinx+2(1-2sin²(x\2))-2sinx(1-2sin²(x\2))-1=0
sinx+2-4sin²(x\2)-2sinx+4sinxsin²(x\2)-1=0
4sinxsin²(x\2)-4sin²(x\2)-sinx+1=0
sinx(4sin²(x\2)-1)-(4sin²(x\2)-1)=0
(4sin²(x\2)-1)(sinx-1)=0
x=pi\2

 

[robbo_145]

hmm
RHS = sin²x +2sinxcosx + cos²x
= (sinx + cosx)²

∴
(sinx + cosx)² - (sinx + cosx) = 0
(sinx + cosx)(sinx + cosx - 1) = 0

sinx + cosx = 0
x = 3π/4 7π/4

or
sinx + cosx = 1
x = 0, π/2, π, 3π/2, 2π

late night solution so ive prolly done something wrong

 

[jake2.0]

robbo_145 is almost right sol:x =0, 3π/4, 7π/4, π/2

 

[香港!]

oo i missed out a x=0 and x=2pi somewhere in there -_-

 

[robbo_145]

sinx + cosx = 1
x = 0, π/2, π, 3π/2, 2π

woops
should just be

x = 0, π/2, 2π

 

[Bellow]

ive got extra answers for some reason....

sinx + cosx = 1 + sin2x
(sinx + cosx)² = (1 + sin2x)²
1 + sin2x = 1 + 2sin2x + sin²2x
0 = sin2x + sin²2x
0 = sin2x(1 + sin2x)

sin2x = 0
2x = 0, π, 2π, 3π, 4π
x = 0, π/2, π, 3π/2, 2π

1 + sin2x = 0
sin2x = -1
2x = 3π/2, 7π/2
x = 3π/4, 7π/4

Therefore, x = 0, π/2, 3π/4, π, 3π/2, 7π/4, 2π

 

[香港!]

i think if u use square then u create extra solutions...

 

[+Po1ntDeXt3r+]

x = 0, pi/2, 3/4 pi, pi, 3/2 pi, 7/4 pi, 2pi

this will generate all plus extras so u sub them in cos its the quickest way.... preferably into
(sin x + cos x)(sin x + cos x - 1) = 0 or any of the pre-squaring equations
as this generates the zeros easier..

then x=pi, 3pi/2 are invaild

solution is x = 0, pi/2, 3pi/4, 7pi/4, 2pi

yeah im just posting for completion sake