Please help solve... (2 Viewers)

[study-freak]

that just says theres no tp's or inflexion.

I see....

No idea how to do this question then...

 

[gurmies]

Lolwut, why differentiate? What do you think, there are multiple roots or something? -.-

 

[dp624]

If anyone can solve..
cos^2 x=tan x
1=tan x sec^2 x
1=tan x (1+tan^2 x)
tan^3 x + tan x -1=0

I just can't solve after this..

let tan x = y
y^3 + y - 1 = 0
and there's some formula which can be used for solving degree-3 equations

but that's sort of cheating, no?

 

[study-freak]

Lolwut, why differentiate? What do you think, there are multiple roots or something? -.-

just tried since I could think of no other ways

 

[study-freak]

let tan x = y
y^3 + y - 1 = 0
and there's some formula which can be used for solving degree-3 equations

but that's sort of cheating, no?

What's the formula?

 

[lyounamu]

What's the formula?

You don't know THE FORMULA???????????????????????????????

 

[davidyin92]

What's the formula?

It's not pretty, square roots within square roots and it's really long: if it is one that is similar to the quadratic formula you are looking for

 

[study-freak]

You don't know THE FORMULA???????????????????????????????

LOL I don't know the formula for 3-degree ones...


Edit: Just discovered by googling 'cubic formula'...

 

[lyounamu]

LOL I don't know the formula for 3-degree ones...

You have to know THE FORMULA if you want to get this question right.

 

[dp624]

yeah it's not pretty
LOL i had a calculator that could give me solutions to 2,3,4 degree roots - but it wasn;t allowed in the exam =S

 

[Drongoski]

Next somebody's gonna pose Fermat's "Last Theorem" on BOS for a quick solution!

 

[gurmies]

lol! Cordano ftw =)

 

[shaon0]

lol! Cordano ftw =)

lol.

Let z=cos(x)
[cos(x)]^2={sqrt(1-[cos(x)]^2)/cos(x)}
z^3=sqrt(1-z^2)
z^6+z^2-1=0
And that's all i can do

 

[ForbiddenND]

If anyone can solve..
cos^2 x=tan x
1=tan x sec^2 x
1=tan x (1+tan^2 x)
tan^3 x + tan x -1=0

I just can't solve after this..

urr can you use quadratics or some other thing like that?

 

[kwabon]

dunno whether right or not

cos^2 x =tan x
cos^2 x = sin x / cos x
cos^3 x= sin x
cos^3 x/sin x = 1
cot x ( cos^2 x) = 1
cot x = 1 / cos^2 x
1/tan x = sec^2 x
1= tan x(tan^2 x +1 )

therefore
tan^2 x + 1 =1
tan^2 x = 0
tan x = 0
x = pi*n
(where n is an integer)

and
tan x = 1
x = pi*n + pi/4
(where n is an integer)

hence
x = pi*n + pi/4 and x = pi*n

lol, thats wat i got, and think its wrong

hope it helped and good luck!!

 

[Timothy.Siu]

dunno whether right or not

cos^2 x =tan x
cos^2 x = sin x / cos x
cos^3 x= sin x
cos^3 x/sin x = 1
cot x ( cos^2 x) = 1
cot x = 1 / cos^2 x
1/tan x = sec^2 x
1= tan x(tan^2 x +1 )

therefore
tan^2 x + 1 =1
tan^2 x = 0
tan x = 0
x = pi*n
(where n is an integer)

and
tan x = 1
x = pi*n + pi/4
(where n is an integer)

hence
x = pi*n + pi/4 and x = pi*n

lol, thats wat i got, and think its wrong

hope it helped and good luck!!

nope
if x^1000 (x+1)=1
can u say therefore x+1=1, x=0 LOL

using your method we ever need to use the quadratic formula
e.g.
x^2+5x+2=0
x(x+5)=-2
x=-2 or x+5=-2
x=-2 or x=-7

ur method would only work, IF the solutions were THE SAME for tanx=1 and tan^2 x=0

 

[clintmyster]

dunno whether right or not

cos^2 x =tan x
cos^2 x = sin x / cos x
cos^3 x= sin x
cos^3 x/sin x = 1
cot x ( cos^2 x) = 1
cot x = 1 / cos^2 x
1/tan x = sec^2 x
1= tan x(tan^2 x +1 )

therefore
tan^2 x + 1 =1
tan^2 x = 0
tan x = 0
x = pi*n
(where n is an integer)

and
tan x = 1
x = pi*n + pi/4
(where n is an integer)

hence
x = pi*n + pi/4 and x = pi*n

lol, thats wat i got, and think its wrong

hope it helped and good luck!!

wouldn't you know straight away its wrong if you sub back your solution into the original equation?

 

[kwabon]

wouldn't you know straight away its wrong if you sub back your solution into the original equation?

LOL, yep its wrong
oh well

 

[zzdfa]

nvm

 

[cutemouse]

If anyone can solve..
cos^2 x=tan x
1=tan x sec^2 x
1=tan x (1+tan^2 x)
tan^3 x + tan x -1=0

I just can't solve after this..

[maths]x=\tan ^{(-1)}\left[-\left(\frac{2}{3 \left(9+\sqrt{93}\right)}\right)^{1/3}+\frac{\left(\frac{1}{2} \left(9+\sqrt{93}\right)\right)^{1/3}}{3^{2/3}}\right]\\=\tan ^{(-1)}\left[\frac{-2 \left(\frac{3}{9+\sqrt{93}}\right)^{1/3}+\left(2 \left(9+\sqrt{93}\right)\right)^{1/3}}{6^{2/3}}\right]\\x=\tan ^{(-1)}\left[-\frac{\left(1+i \sqrt{3}\right) \left(\frac{1}{2} \left(9+\sqrt{93}\right)\right)^{1/3}}{2 3^{2/3}}+\frac{1-i \sqrt{3}}{2^{2/3} \left(3 \left(9+\sqrt{93}\right)\right)^{1/3}}\right]\\x=\tan ^{(-1)}\left[-\frac{\left(1-i \sqrt{3}\right) \left(\frac{1}{2} \left(9+\sqrt{93}\right)\right)^{1/3}}{2 3^{2/3}}+\frac{1+i \sqrt{3}}{2^{2/3} \left(3 \left(9+\sqrt{93}\right)\right)^{1/3}}\right][/maths]

EDIT: How do I do line breaks in LateX :S -- Thanks Jetblack