Question

can any one help me please.....
what are the roots of (sinx)^5+(cosx)^3=1

Answer 1

2 roots are x=0, x=pi/2

Answer 2

please show your work.....thnks

Answer 3

to get the rest you have to solve an 8th degree polynomial using approximation methods

Answer 4

ok..how do you start them please?

Answer 5

well think what makes sinx =0 and cosx=1
this yields x=0
well think what makes sinx =1 and cosx=0
this yields x=pi/2

Answer 6

yeah correct,.... what about the polynomial?

Answer 7

ok i assigned u=sinx and v=cosx
u=sinx = sqrt(1-cosx^2) = sqrt(1-v^2)
from u^5 +v^3=1
u = 5th root of (1-v^3)
set these equal to each other and raising to power of 10 yields
(1-v^2)^5 = (1-v^3)^2
=> v^2(v^8-5v^6+11v^4-10v^2-2v+5) = 0

Answer 8

wow thank you so much...thats very good....

Answer 9

your welcome

Answer 10

you are really the life saver here....lol thnk you so much again......

Answer 11

:)
also depending on the directions, pi would also work

Answer 12

wait nevermind pi does not work
cospi = -1

Answer 13

ah ok......

Answer 14

can we try also euler identities here?

Answer 15

ok thnks you so much again

Answer 16

umm maybe, usually thats when you are dealing with complex numbers isn't it
im not an expert on eulers

Answer 17

ok for euler formula
cosx=(e^ix + e^-ix)/2
sinx=(e^ix - e^-ix)/2i

Answer 18

ok if we use that we still get a messy equation
(e^ix - e^-ix)^5 + 4i(e^ix +e^-ix)^3 = 32i

Answer 19

thats true its too messy...lol so ill stick to polynomials lol....

Answer 20

thanks again for you great great help......

Answer 21

no problem