whats the answer to sin(2x)+1= Cos(2x)?

Answer 1

All the values of X which satisfy the equation cos(2x)-sin(2x)=1. Say for example n*pi. where
n=............-1,0,1,............
similarly x=-(n+1)*pi/2 where n=0,1,2,.......... no negative values in this case.
It can have many cuch solutions.

Answer 2

Let´s use tan x
2 tan x/(1+tan^2x)+1= (1-tan ^2x)/(1+tan^2x) call tan x =t
2t+1+t^2 = 1-t^2
so
2t^2+2t =0 t= 0 so x= k*pi
t=-1 x= -pi/4+k*pi

Answer 3

So, let's rearrange this to get

sin(2x)-cos(2x)=-1

Square both sides to get

sin^2(2x)-2sin(2x)cos(2x)+cos^2(2x)=1

Therefore,

-2sin(2x)cos(2x)=0

And thus we have either

sin(2x)=0 or cos(2x)=0

The first solves as x=n*pi/2
The second solves as x=pi/4 + n*pi/2

However, because we took a square root, we must remember to qualify this by taking out all answers of form x=pi/4 + n*pi

and

x=pi/2 + n*pi

The final answer is x = 3pi/4 + n*pi, n*pi
=)

Answer 4

sin2x + 1 = cos2x
=> sin2x + 1 -- cos2x = 0
=> 2sinxcosx + 2sin^2x = 0
=> sinx {cosx + sinx} = 0
when sinx = 0 = sin0 then
x = nπ. Answer.
when cosx + sinx = 0
=> sin(x + π/4) = 0 = sin0 then
x = nπ -- π/4) Answer.