Question

Please help! How do you solve for x?
4sin(x)=cos(x)-2

Answer 1

let y=sin(x), sqrt(1-y^2)=cos(x). solve for y, then x=arcsin(y)+2*pi*C

Answer 2

How did you come to that?

Answer 3

since it is difficult to solve for x, solve for sin(x) instead. then x=arcsin(sin(x))+2*pi*C

Answer 4

I have not learned arcsin yet. Is there a way to solve this problem without arcsin and maybe patharoyean identities?

Answer 5

using the fact that sin^2+cos^2=1, you can reduce the equation down to
4 sin=sqrt(1-sin^2)-2
(4 sin+2)=sqrt(1-sin^2)
16 sin^2+16 sin+4=1-sin^2
17 sin^2+16sin+3=0
sin x=(sqrt(13)-8)/17
beyond this point, arcsin is necessary to find x

Answer 6

In the first line, How did you get the expression under the radical since it is cos, not \[\cos^{2}\]? I attached what i've done, but when I use the inverse cos button on my calc, it says doamin error and I dont know why it wont work.

Answer 7

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

Answer 8

(cos(x)-2)^2=cos^2+4-4cos(x)

Answer 9

i think you missed a term

Answer 10

okay. I am going to try and redo it with what you suggested. What do you think the answer is though?

Answer 11

in radians or in degrees?
in radians, -0.261+2*pi*n where n is an integer
in degrees, -14.981+360*n, where n is an integer

arcsin((sqrt(13)-8)/17)