C1 and C2 DOUBTS HERE!!!!!

[astarmathsandphysics]

4 sin x = tan x
4 sin x = sin x/cos x
Sin x (4 -1/cos x)=0
Sin x =0 so x =0,180,360
4-1/cos x =0 so cos x =1/4 so x =cos^-1 1/4 or 180-cos^-1 1/4

[~ Miss Relina ~]

4 sin x = tan x
4 sin x = sin x/cos x
Sin x (4 -1/cos x)=0
Sin x =0 so x =0,180,360
4-1/cos x =0 so cos x =1/4 so x =cos^-1 1/4 or 180-cos^-1 1/4

Thanks alot Sir for your support 

[astarmathsandphysics]

no probs

[~ Miss Relina ~]

another question to bother this thread about is here

for who have C2 book its on page 193 Q 6 in the revision exercise 3

what I don't understand is that the book's answer is 4 and when I did integration of the two parts of my  sketch , one area was -ve and other was +ve so I added them ignoring the -ve sign so...............are these steps right or I've just missed somethnig up cause my anwser was ''9.5''

Thanks in advance 

[Arthur Bon Zavi]

Integration gives you :

x⁴    
--- - x³
4

[(x⁴/4) - x³]⁴₂

[{((256/4) - 64) - (4)³) - {(16/4) - (2)³}]⁴₂

| [(0) - (-4)] | = 4

[~ Miss Relina ~]

Integration gives you :

x⁴    
--- - x³
4

[(x⁴/4) - x³]⁴₂

[{((256/4) - 64) - (4)³) - {(16/4) - (2)³}]⁴₂

| [(0) - (-4)] | = 4

Thanks
but have you done a sketch as when I've done a sketch the I found two parts above and under the curve that's why I used two intergration formulas one with upperlimit 4 and lowerlimt 3 (where the curve intersect the x-axis) and other intergration with the upperlimit 3 and lowerlimit 2 but i got it wrong ...........so my sketch may be wrong or what

anyways Thanks you very much

[Arthur Bon Zavi]

Thanks
but have you done a sketch as when I've done a sketch the I found two parts above and under the curve that's why I used two intergration formulas one with upperlimit 4 and lowerlimt 3 (where the curve intersect the x-axis) and other intergration with the upperlimit 3 and lowerlimit 2 but i got it wrong ...........so my sketch may be wrong or what

anyways Thanks you very much

Yes, your sketch may be incorrect.

[~ Miss Relina ~]

Yes, your sketch may be incorrect.

but I think it may be right sorry for disturbing you but I'll write the steps if anything wrong tell me

x³-3x²=0
x²(x-3)=0
thus x = 0 or 3
so it may look like that attached

[astarmathsandphysics]

Exactly right.

[~ Miss Relina ~]

Exactly right.

then what's wrong with  that integration question what is my mistake ? I 'm really confused
anyways Thanks Sir

[astarmathsandphysics]

The graph is part below and part above the x axis.
Integrate between 2 and 3. This will be negative because it is below the x axis, so make it positive.
Integrate between 3 and 4. This will be positive.
Add the two areas to give the answer.

[~ Miss Relina ~]

The graph is part below and part above the x axis.
Integrate between 2 and 3. This will be negative because it is below the x axis, so make it positive.
Integrate between 3 and 4. This will be positive.
Add the two areas to give the answer.

I did so but my answer turned to be wrong 
and when Arthur did it with integration as a whole of upperlimit 4 and lowerlimit 2 it gave the book answer and that's what I'm confused about

[astarmathsandphysics]

I will do a full answer When I get home

[astarmathsandphysics]

Here it is

[~ Miss Relina ~]

Here it is

exactly the same as my answer  

but the problem is that the book's answer is 4  may be its wrong  

Integration gives you :

x4    
--- - x3
4

[(x4/4) - x3]42

[{((256/4) - 64) - (4)3) - {(16/4) - (2)3}]42

| [(0) - (-4)] | = 4



that's Arthur answer  

so what's your opinion Sir may be that book is wrong