Question

solve the easy(trig)nometric equation

Answer 1

\[\sin^3x(1+\cot x)+\cos^3(1+\tan x)=\cos 2x\]

Answer 2

write cos 2x = cos^2x-sin^2x = (cos x-sin x)(cos x+ sin x)
u also get the factor cos x + sin x on left side, u know how ?

Answer 3

\[\sin^3 x+\cos^3x+ \sin^2x \cos x + \cos^2x \sin x=\cos 2x\]
opening gives

Answer 4

yes
\[(\sin x+\cos x)(1-\sin x \cos x)+\sin x cosx(\sin x+\cos x) \]

Answer 5

\[(\sin x+\cos x)(1-\sin x \cos x +\sin x \cos x)=LHS\]

Answer 6

don't do that ...
s^2(s+c)+c^2(s+c) = (s^2+c^2)(s+c) = 1(s+c) = s+c = LHS

Answer 7

yeah, u also get same

Answer 8

oh thanks so
\[s+c=s^2-c^2\]

Answer 9

\[s=c,\tan x=1\]

Answer 10

so, u get 1=s-c

Answer 11

how s=c ?

Answer 12

oh i cancelled which is not allowed

Answer 13

\[c+s=(s-c)(s+c)\]

Answer 14

from that ^ , u get
s+c=0 or s-c=1

Answer 15

yes

Answer 16

can u solve those 2 ?

Answer 17

yes add them
s=1,c=1/2

Answer 18

seperatly

Answer 19

not simulatneously!

Answer 20

s+c=0
get values of x from here

Answer 21

\[\tan x=-1\]
2nd 1 weird

Answer 22

s-c=1
(1/sqrt2)s - (1/sqrt2)c = 1/sqrt

Answer 23

\[\sin(\pi/4+x)=1\]

Answer 24

(1/sqrt2)s - (1/sqrt2)c = 1/sqrt2
cos 45 sin x - sin 45 cos x = sin 45 or cos 45

Answer 25

\[\sin(\pi/4-x)=1\]

Answer 26

=1 ?

Answer 27

= sin 45

Answer 28

sorry 1/sqrt2

Answer 29

so now we can combine the two

Answer 30

\(\sin(\pi/4-x)=\sin (\pi/4)\)

Answer 31

x= 0,2pi,4pi,.....2n pi

Answer 32

yes i like the way you use c and s its way better than the whle thing esp with long eq

Answer 33

and from s+c=0
what are values of x ?

Answer 34

2nd sol x=0+2 pik,
x=pi+2pi k

Answer 35

0,pi,2pi,3pi....

Answer 36

npi

Answer 37

i am getting hungry

Answer 38

tan (n*pi) is not -1

Answer 39

i was doing the wrng one wait

Answer 40

\[x=\frac{ 3 }{ 4 }\pi+2\pi k\]
\[(3/4\pi,7/4\pi,11/4\pi,....(2n+1)/4)\]

Answer 41

\[\pi k\]not 2pi k

Answer 42

are we there yet

Answer 43

yeah, now its correct
3pi/4+pi*k

Answer 44

great...thanks

Answer 45

welcome ^_^

Answer 46

can i post last 1

Answer 47

sure, but i m hungry too :P

Answer 48

lol,that makes 2 of us

Answer 49

i dont get this how do u do it?? is this proving both side equal do you start left side

Answer 50

this is called general solution,we are solving for x ,not proving

Answer 51

hey what do you do after the opening gives: part.. factor out?

Answer 52

yes sin x+ cos x

Answer 53

how do you get s+c=0? and how did jonask factor s+c at same time? or was it wrong

Answer 54

c+s=(s−c)(s+c)
u asking how from here s+c= 0 ?

Answer 55

yes

Answer 56

c+s=(s−c)(s+c)
0=(s−c)(s+c)-(s+c)
0=(s+c)(s-c-1)
so
s+c=0
or
s-c-1=0

Answer 57

ohh i just divided and only got s-c=1

Answer 58

i thinkskipping steps is not wise sometimes cos you become erronous

Answer 59

does dividing usually result in missing solns?

Answer 60

and how did you factor s^3+c^3 into (s+c)(1-sc)

Answer 61

u know s^2+c^2=1 ?

Answer 62

yeah

Answer 63

and a^3+b^3 formula ?

Answer 64

perfect cube? nope ..

Answer 65

is that formula from binomeal theorem or something

Answer 66

\(\huge a^3+b^3=(a+b)(a^2-ab+b^2)\)

Answer 67

ah got it.. thanks.. and the rest is alright.. we havent done general solns though