Question

how would you solve this?
2x+cosx=1

Answer 1

i get that if x=0 than cos(0)=1 and 2(0)=0 so x should be 0 but is there another way of solving it?

Answer 2

This is a great question and unfortunately there's not a good answer. You could consider graphing 2x+cosx and seeing where it intersects the line y=1.

Your intuition should consider that cos(x) is bounded from 1 and -1 so you can narrow down the range you need to graph by looking at the extreme cases of when you could possibly have an answer:

2x+1=1
and
2x-1=1

So solving for x in these cases we have

x=0 and x=1 as the only possible range an answer can even lie, which is pretty good.

Answer 3

thats a bummer, but thanks for explaining it

Answer 4

cosx=1-2x

Answer 5

\[-1 \le cosx \le1\]

Answer 6

\[-1 \le 1-2x \le 1\]

Answer 7

subtract 1 from each

Answer 8

\[-2 \le -2x \le0\]

Answer 9

Here's just a fun example that you might find interesting or not, I don't know haha but I think it's interesting.

For instance, let's suppose we had a different trig function instead. Noting that cos(x)=sin(x+pi/2) let's just throw in an arbitrary phase angle and see what we would need for the answer to fall at x=1.

\[\LARGE 2x+\sin(x+\phi)=1\] So we know trig functions can only have a max and min of -1 and 1 like we described, and so this will lie on the same interval from 0 to 1 for any angle. Since cosine satisfies the end when x=0, we plug in x=1 to find out what trig function will satisfy the other end purely for fun! \[\LARGE 2+\sin(1+\phi)=1\] rearranging and solving we get: \[\LARGE \phi = \frac{3\pi}{2}-1\] so the other function is \[\LARGE 2x+\sin(x+\frac{3\pi}{2}-1)=1\] when x=1 is the single solution.

Any other answer you want can probably easily be found as long as you allow sin(x+phi)=1-2x to be easily solvable. Anyways haha.

Answer 10

divide each by -2

Answer 11

\[0 \le x \le 1\]

Answer 12

thanks @gorv thats makes sense! but than wouldn't x be 1 or 0

Answer 13

ohm right u just test it with which ever one works! thanks a lot!

Answer 14

thanks guys @gorv @Kainui

Answer 15

yeah but there will be many value which satisfy ..and will be between 0 and 1 including 0 and 1

Answer 16

wlcm :)