Trig!!!!! (1 Viewer)

[atakach99]

solve for 0< (or equal to) X < (or equal to) 2pie

cos squared x - sin x = -1

 

[webby234]

cos²x - sin x = -1
1 - sin²x - sinx = -1
sin²x + sinx - 2 = 0
(sinx - 1)(sinx + 2) = 0
So sin x = 1 (as sinx can't equal -2)
So x = pi/2

 

[tacogym27101990]

回复: Trig!!!!!

cos²x - sinx = -1
(1 - sin²x) - sinx = -1
-sin²x - sinx +2=0
sin²x + sinx - 2=0
(sinx+2)(sinx-1)=0
sinx=-2, sinx= 1
no solutions for the first one, x=pi/2 for the second one
therefore x=2

 

[conics2008]

Re: 回复: Trig!!!!!

looks like people already of done it.. think logic not maths.

 

[conics2008]

Re: 回复: Trig!!!!!

looks like people already of done it.. think logic not maths.

 

[atakach99]

well how would u do this one

cos x + square root 3 sin x =2

 

[conics2008]

well how would u do this one

cos x + square root 3 sin x =2

use transformation method, its the best one out, or if your good at theta/2 method use it. ill give u the method.

R= square root of 3+1 = 4 = 2

tan a = root of 3 over 1 = 60 = pi/3

therefore

2cos(x-60) = 2

divide by 2 and then take cos of 1 = 0

so x - 60 = 0 and also 360

therefore x = 420
you cant really use -60

solutions are 60,420, 780 .... 2pin + 60 =)

according to my head where n is an integer and number of revolution.

 

[atakach99]

sorry man but i didnt understand how u got that answer
did u solve it using the t- rules???

 

[conics2008]

sorry man but i didnt understand how u got that answer
did u solve it using the t- rules???

u

naa i used transformation method...

it is used to solve any trig equation in the form of

acosx (+ - ) b sin x = C
or
bsinx (+-) a cos x = C

if you want me to use the t crap, ill post it up...

look in your text book or notes for transformation, or do you want me to explain to you how to use transformation method ??

 

[atakach99]

oh ok i will need to revise the transformation method
but if u dont mind can u solve it with the "t - crap"

 

[lyounamu]

All the necessary working out is like the following:

Let cos (x) + square root 3 sin (x) = rcos (x-a)
= r(cos (x)cos(a) + sin (x)sin(a))
By equating we get:

1= rcos(a) - (eq. 1) and square root 3= rsin(a) - (eq. 2)

Now, divide the (2) by (1) and we get: tan(a) = squar root 3/1
Therefore, a = pi/3

Substitute the value into either (1) or (2) to find the value of r.
Therefore, r = 2

Then, cos(x) + square root 3 sin(x) = 2cos(x-pi/3)
Therefore, 2cos (x-pi/3) = 2
cos (x-pi/3) = 1
x-pi/3 = 2pi +- (+ and -) + 0
x = 2piN + pi/3 (general solution)

 

[conics2008]

This one follows the t formula.. i made a mistake with my sign, you can use 60 but not -60.. sorry about that.

you know sin x/2 = 2t/1+t^2 and cos x/2 = 1-t^2/ 1+t^2 right ??

sub them in you get

1-t^2/1+t^2 + 2root3 t / 1+t^2 = 2

simpify you get ( that is remove deno )

1-t^2 +2root3 t = 2 + 2t^2

3t^2- 2root3 t +1 = 0

use quadratic

at the end you get t = 1/root3 after rationalising dont use root3/3 its same thing but when u rationalise it, its more easier ( its basicallly the same )

therefore t= tan x/2 >>>

tan x/2 = 1/root3 >> x/2 = 30 therefore x = 60 and then use your trig knowledge to find the other angels. =)

 

[conics2008]

All the necessary working out is like the following:

Let cos (x) + square root 3 sin (x) = rcos (x-a)
= r(cos (x)cos(a) + sin (x)sin(a))
By equating we get:

1= rcos(a) - (eq. 1) and square root 3= rsin(a) - (eq. 2)

Now, divide the (2) by (1) and we get: tan(a) = squar root 3/1
Therefore, a = pi/3

Substitute the value into either (1) or (2) to find the value of r.
Therefore, r = 2

Then, cos(x) + square root 3 sin(x) = 2cos(x-pi/3)
Therefore, 2cos (x-pi/3) = 2
cos (x-pi/3) = 1
x-pi/3 = 2pi +- (+ and -) + 0
x = 2pi + pi/3 (general solution)

just need to put N in =)

 

[lyounamu]

oh ok i will need to revise the transformation method
but if u dont mind can u solve it with the "t - crap"

Ok.

Let t=tan (x/2)

Therefore, cos x = (1-t^2)/(1+t^2)
sin x = 2t/(1+t^2)

Now, (1-t^2)/(1+t^2) + square root 3 . (2t/(1+t^2) = 2

Then (-t^2 + 1 + square root 3 . 2t)/(1+t^2) = 2
And -t^2 + 1 +square root 3 . 2t = 2 + 2t^2
And 3t^2 - 2 . square root 3 . t +1 = 0

By using quadratic formula we can find that t = 1/ square root 3.
Therefore tan (x/2) = 1 / square root 3
x/2 = pi / 6
x = pi /3

Cannot be bothered to show another general solution one. Sorry! (look above solution I did).

You are also suppoed to test all the points like 0, pi but I didn't do it because... (it's too much effort... I am sorry)

 

[lyounamu]

just need to put N in =)

Thanks!

 

[conics2008]

Thanks!

xD

 

[atakach99]

ok thnks guys now i get...

looks like u guys dont use the t-crap as much as me
for some reason i find it easier but anyway thnks a lot

 

[conics2008]

ok thnks guys now i get...

looks like u guys dont use the t-crap as much as me
for some reason i find it easier but anyway thnks a lot

cuz if you use the t crap your most likely to make mistaakes because t pretty much looks like a plus sign + ... get the picture.

and its way longer, i would strongly recommend you to use the transoformation crap =) its alot better =)

anywho good luck=0

 

[lyounamu]

ok thnks guys now i get...

looks like u guys dont use the t-crap as much as me
for some reason i find it easier but anyway thnks a lot

The reason why I don't use t-formula much is that you have to test some points such as 0 and pi as they may satisfy the equation. I just use the transformation as I am more familiar with that. But it's entirely up to you.

By the way, I sometimes use t-formula when trying to prove that L.H.S = R.H.S.