Core2 Trig - Require Method

Jan 2007 Core2 Paper

"Find all the solutions, in the interval 0 <= x < 2pi, of the equation

2cos^2(x) + 1 = 5 sin(x)

giving each solution in terms of pi"

I've done the whole C2 paper except this question, and gone back to it a few times.

I'm just requesting a full workout and method for this so if I come across another one I will know how to tackle it, thanks

Jorge x

Reply 1

Evil Monkey

I can't do Latex too well, so it'd be really ugly if i tried, so i'll just say, you can swap cos^2 for 1-sin^2, and you've got a quadratic for sin. They should all go something like that.

Reply 2

Mathematix

I've tried, the question is just not working for me :frown:

Reply 3

vc94

Original post by Jorgeyy

I've tried, the question is just not working for me :frown:

When you replaced cos^2 with 1-sin^2 what quadratic equation did you get?

Reply 4

Mathematix

Original post by vc94

When you replaced cos^2 with 1-sin^2 what quadratic equation did you get?

I'll find my workings in my book later,

but trust me, ive left this alone for 24h come back to it and im still stuck

I would appreciate if someone could organise the answer in steps so I can see how to tackle these questions :smile:

Reply 5

sohail.s

Original post by Jorgeyy

Jan 2007 Core2 Paper

"Find all the solutions, in the interval 0 <= x < 2pi, of the equation

2cos^2(x) + 1 = 5 sin(x)

giving each solution in terms of pi"

Original post by Jorgeyy

I would appreciate if someone could organise the answer in steps so I can see how to tackle these questions :smile:

Right.

2cos^2x + 1 = 5sinx
(You MUST remember this identity : cos^2x + sin^2x = 1 which implies that cos^2x = 1 - sin^2x )
Substitute this into your equation (pretty much like solving a simultaneous equation)

cos^2x=1-sin^2x

2 (1 - sin^2x) + 1 = 5sinx

2 - 2sin^2x + 1 = 5sinx

rearrange to get it in the form of a quadratic.

2sin^2x + 5sinx - 3 = 0
It might look scary but it's exactly like a quadratic you solved in Core 1.

TO make it more familiar i'll replace sinx with y .

so 2y^2 + 5y - 3 =0

factorising : (2y-1) ( y+3)=0

Since sinx has a maximum value of 1 and a minimum value of -1 the solution of y=-3 would be rejected (radians or degrees, doesnt matter)

so our solution is that 2y-1 = 0
therefore y = 1/2

so back to our question sinx = 1/2

to find your first solution just do sin^-1 on your calculator (radian mode as it asks in terms of pi)

SO our first solution is sin^-1(1/2) = pi/6

The second solution you would use the CAST method and find that sin is positive in the first two quadrants only.

So the second solution is 180 (pi) - pi/6 which = 5pi/6

And that's all. pretty standard question, just make sure you know the identity :biggrin:

.

Also if in doubt substitute your solutions into the starting equation and it should work ( if not go back and check your solution)

(edited 13 years ago)

Reply 6

Mathematix

Original post by sohail.s

Right.

2cos^2x + 1 = 5sinx
(You MUST remember this identity : cos^2x + sin^2x = 1 which implies that cos^2x = 1 - sin^2x )
Substitute this into your equation (pretty much like solving a simultaneous equation)

cos^2x=1-sin^2x

2 (1 - sin^2x) + 1 = 5sinx

2 - 2sin^2x + 1 = 5sinx

rearrange to get it in the form of a quadratic.

2sin^2x + 5sinx - 3 = 0
It might look scary but it's exactly like a quadratic you solved in Core 1.

TO make it more familiar ill replace sinx with y .

so 2y^2 + 5y - 3 =0

factorising : (2y-1) ( y+3)=0

Since sinx has a maximum value of 1 and a minimum value of -1 the solution of y=-3 would be rejected (radians or degrees, doesnt matter)

so our solution is that 2y-1 = 0
therefore y = 1/2

so back to our question sinx = 1/2

to find your first solution just do sin^-1 on your calculator (radian mode as it asks in terms of pi)

SO our first solution is sin^-1(1/2) = pi/6

The second solution you would use the CAST method and find that sin is positive in the first two quadrants only.

So the second solution is 180 (pi) - pi/6 which = 5pi/6

And that's all. pretty standard question, just make sure you know the identity :biggrin:

.

Also if in doubt substitute your solutions into the starting equation and it should work ( if not go back and check your solution)

I love you.

I forgot the limits to the Sin and Cos graphs of -1 to 1 and that messed me over lol !

Also, is it ok to use the formula?

Sin being... 180 - angle1
Cos being... -angle1
Tan being... angle1 + 180

I should really learn the CAST method properly as well but i'm more of a formula person :biggrin:

Reply 7

sohail.s

Original post by Jorgeyy

yeah thts fine tbh, all of them originate from the cast method tbh

so sin = 180 - angle1
cos = 360 - angle1
tan = 180 + angle 1

Reply 8

Mathematix

I got taught Cos was -angle1 (+n360) so same thing I guess just teachers preference :P

Thanks again man, we've just started covering trig in class its our last core2 topic so hopefully I'll understand it lightning speed ready for the exam!

Big rep x

Reply 9

Americaniamh

Original post by Jorgeyy

Is Core 2 part of A level Maths or F. Maths?

Reply 10

Mathematix

Original post by Americaniamh

Is Core 2 part of A level Maths or F. Maths?

Core1 + Core2 + a side module such as stats or mechanics etc makes up AS Maths :smile:

Reply 11

Americaniamh

Original post by Jorgeyy

Core1 + Core2 + a side module such as stats or mechanics etc makes up AS Maths :smile:

Thanks!

Reply 12

Tink.0818

2cos^2x+1=5sinx
Replace cos^2x for 1-sin^2x
2(1-sin^2x)+1=5sinx
2-2sin^2x+1=5sinx
0=2sin^2x+5sinx-3
Replace sinx with y
2y^2+5y-3=0
Factorise
(2y-1)(y+3)
Sinx=1/2 sin-1x=30
I can get this far but don't understand how to give your answers in terms of pi...if anyone can help

Reply 13

Parallex

Original post by Tink.0818

jesus I thought you were answering a 4 year old question then..

sin x = 1/2
arcsin (1/2) = 30 degrees

there's other answers as well, 150 for example

to get from degrees to radians, divide by 180 then multiply by pi

so 30 degrees in radians is 1/6pi etc