Question

how do i solve for c in the equation 2sinc(1-cosc) = 4/pi ?

Answer 1

hmmm can you post a picture of the material? I'm assuming that might be a typo =)

Answer 2

i typed it correctly but i'll post a pic

Answer 3

\[2[\sin(c)] - \sin(2c) = 4/\pi\]

Answer 4

then i changed sin(2c) using the double angle formula

Answer 5

I see

Answer 6

\[2\sin(c) - 2\sin(c)\cos(c)\] = 4/pi

Answer 7

then i factored out 2sinc

Answer 8

now i do not know how to solve for c

Answer 9

if you factored and divided by the constant you may want to take the reciprocal. that way you at least have pi/something instead of something over pi. may be a bit more comprehendable.

Answer 10

take the reciprocal of what?

Answer 11

\[2\sin(c) \times (1-cosc) = 4/\pi\]

Answer 12

\[sinc \times (1-cosc) = 2/\pi\]

Answer 13

nvm... you can sub sin^2 + cos^2 for 1

Answer 14

wait a sec...

Answer 15

i'll wait

Answer 16

drawing a blank... what class is this for?

Answer 17

calculus. a blank?

Answer 18

is this an area or something? what was the original question/setting?

Answer 19

or a derivative? (by area i meant integral)

Answer 20

drawing a blank = no good ideas

Answer 21

so what was the original question that this came from?