Question

Find all solutions of (sinx)^5+(cosx)^3=1.
Show ur work. or just explain the process.

Answer 1

you can use euler formula for sin and cos

Answer 2

\[\sin ^{5}x =1-\cos ^{3}x\]

Answer 3

wat after tat mark o?

Answer 4

cos x=(e^ix +e^-ix)/2 , sin x= (e^ix -e^-ix)/2i

Answer 5

\[\sin ^{10}x =1-2\cos ^{3}x +\cos ^{6}x\]

Answer 6

mark o:wat do after tat?
@chaguanas: please proceed. :)

Answer 7

\[(1-\cos ^{2}x)^{5}=1-2\cos ^{3}x +\cos ^{6}x\]

Answer 8

Everything is in cos. Finish up.

Answer 9

ya...i ended up in a 6degree polynomial...then how to proceed?

Answer 10

can u show the whole sum..dont leave midway please.. :)

Answer 11

the euler formulas of sin and cos that i gave you expand them for (sinx)^5+(cosx)^3=1.

Answer 12

then what?

Answer 13

then youl end up with sin and cos answers

Answer 14

can u xplain more or show some more steps...mark o?

Answer 15

please help..its urgent!

Answer 16

\[1-4\cos ^{2}x +5\cos ^{4}x -2\cos ^{6}x +\cos ^{8}x -\cos ^{10}x =1-2\cos ^{3}x +\cos ^{6}x\]

Answer 17

ya..then u cancel one on both sides and get cosx=0 as repeated root. Also u get (cosx-1)^2 as a factor. So ur left with 10-4=6 roots.What to do now chaguanas?

Answer 18

or use (sinx)^5=(10sinx-5sin3x+sin5x)/16
(cosx)^3=(3cosx+cos3x)/4

Answer 19

the maximum value of sine or cosine (for real numbers) is 1. so either you have sinx=1 and cosx=0 (one answer is x=pi/2) or you have sinx=0 and cosx=1 (another answer is x=0).

Answer 20

mark o..ur method will then be similar to chaguanas...my problem is i reduced the problem to a 6 degree polynomial y^6+2y^5+3y^4+4y^3+6y^2+6y+3 where y=sinx. Now how to find roots of this polynomial?

Answer 21

synthetic division

Answer 22

yup synthetic division to get the roots

Answer 23

wat u mean? what is the divisor?

Answer 24

descartes rule of signs shows there there will be atmost 6 negative roots. no positive roots.

Answer 25

u need to know at least one factor for synthetic division! how will u find that?

Answer 26

http://www.purplemath.com/modules/synthdiv.htm

Answer 27

i know synthetic division chaguanas... :) problem is u need a divisor for any kind of division..and u dont know the divisor here!

Answer 28

You try (y+1), or (y-1), trial and error.

Answer 29

not getting thru trials :(

Answer 30

you can graph them to get the ist x1....

Answer 31

use a graphing softwares

Answer 32

:) unless you're allowing x to be a complex number, note that sinx and cosx should be between -1 and 1.

Answer 33

i graphed the beast and found all roots to be complex...so no more real roots....but in my exam we are not allowed any kind of graphing device...is there any way to know that such a polynomial has all complex roots?

Answer 34

If you work this meticulously as i have done on paper, it ends up in a solveable thing\[\cos ^{8}x(1-\cos ^{2}x)=0\]

Answer 35

how did u get that chaguanas?

Answer 36

From my work up top. Put everything on one side. Keep pulling out the factors of cosine.

Answer 37

its not coming can us show chaguanas?

Answer 38

if you use euler formulas
cos x=(e^ix +e^-ix)/2 , sin x= (e^ix -e^-ix)/2i
(cos x=(e^ix +e^-ix)/2)^3 + (sin x= (e^ix -e^-ix)/2i)5

[i(e^-ix - e^ix)^5]/32 +[(e^-ix +e^)^3]/8=1

complex numbers

Answer 39

i mean
(cos x=(e^ix +e^-ix)/2)^3 + (sin x= (e^ix -e^-ix)/2i)5 =1

Answer 40

\[\cos ^{8}x -\cos ^{10}x -3\cos ^{6}x +5\cos ^{4}x +2\cos ^{3}x -4\cos ^{2}x =0\]

Answer 41

so chaguanas solution needs descartes rule of sign and use of synthetic divisions by trials

Answer 42

\[(\cos ^{2}x)(\cos ^{6}x -\cos ^{8}x -3\cos ^{4}x +5\cos ^{2}x +2\cos x -4)=0\]

Answer 43

ok i see that you are not getting thru trials..lol

Answer 44

mark o..can u proceed with ur euler expansion..i want to see ur process!

Answer 45

Keep pulling the factors of cosine out of second bracket.

Answer 46

mark o..my polynomial had no real roots

Answer 47

chaguanas...will u show the next step please!

Answer 48

\[(\cos ^{3}x)(\cos ^{5}x -\cos ^{7}x -3\cos ^{3}x +5\cos x -2)=0\]

Answer 49

Keep pulling out factors of cos. Good luck. Sleep calls.

Answer 50

good luck saubhik,...my pc keeps on hanging up now with me...thnx..

Answer 51

thnx everybody :)