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i got stck on thin trigonometric sum, can anybody help me please?

sum: 5cosxsin2x+4sinsquaredx=4

sum: 5cosxsin2x+4sinsquaredx=4

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#2

(Original post by

i got stck on thin trigonometric sum, can anybody help me please?

sum: 5cosxsin2x+4sinsquaredx=4

**islandguy**)i got stck on thin trigonometric sum, can anybody help me please?

sum: 5cosxsin2x+4sinsquaredx=4

since you didnt give me the range...ill simplify the equation till its solvable by you

5CosXSin2X + 4Sin²X = 4

5CosX(2SinXCosX) + 4(1-Cos²X) = 4

10SinXCos²X + 4 - 4Cos²X = 4

10SinXCos²X - 4Cos²X = 0

(divide the equattion by Cos²X)

10SinX - 4 = 0

SinX = 4/10

X = 23.6, 156.4, 203.6, 336.4

i guess thats the way of doing it at condition that 0<X<360

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#4

(Original post by

ekeke dunno if im right but

since you didnt give me the range...ill simplify the equation till its solvable by you

5CosXSin2X + 4Sin²X = 4

5CosX(2SinXCosX) + 4(1-Cos²X) = 4

10SinXCos²X + 4 - 4Cos²X = 4

10SinXCos²X - 4Cos²X = 0

10SinX - 4 = 0

SinX = 4/10

X = 23.6, 156.4, 203.6, 336.4

i guess thats the way of doing it at condition that 0<X<360

**MalaysianDude**)ekeke dunno if im right but

since you didnt give me the range...ill simplify the equation till its solvable by you

5CosXSin2X + 4Sin²X = 4

5CosX(2SinXCosX) + 4(1-Cos²X) = 4

10SinXCos²X + 4 - 4Cos²X = 4

10SinXCos²X - 4Cos²X = 0

**(divide the equattion by Cos²X)**10SinX - 4 = 0

SinX = 4/10

X = 23.6, 156.4, 203.6, 336.4

i guess thats the way of doing it at condition that 0<X<360

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Determine the values of x and z if:

3lnz - 4lnx = ln2 and 9^(z-2x) = 3^x

I thank u in advance

3lnz - 4lnx = ln2 and 9^(z-2x) = 3^x

I thank u in advance

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#6

(Original post by

Determine the values of x and z if:

3lnz - 4lnx = ln2 and 9^(z-2x) = 3^x

I thank u in advance

**islandguy**)Determine the values of x and z if:

3lnz - 4lnx = ln2 and 9^(z-2x) = 3^x

I thank u in advance

(z^4)/x^4 = 2 (using ln A - ln B = ln A/B and nln A = ln A^n, then exponentiating both sides)

9^(z - 2x) = 3^x

(3^2)^(z - 2x) = 3^x

Bases the same, so powers must be the same

2(z - 2x) = x

Solve those simultaneously.

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#7

**islandguy**)

Determine the values of x and z if:

3lnz - 4lnx = ln2 and 9^(z-2x) = 3^x

I thank u in advance

The second one is (i think)

9^(z-2x) = 3^x

3^2(z-2x)=3^x

so x is

2(z-2z)=x

z is

2z-4z=x

-2z=x

z=x/-2

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#8

(Original post by

wot level is this? i dont know what the first one is

The second one is (i think)

9^(z-2x) = 3^x

3^2(z-2x)=3^x

so x is

2(z-2z)=x

z is

2z-4z=x

-2z=x

z=x/-2

**Tazzie**)wot level is this? i dont know what the first one is

The second one is (i think)

9^(z-2x) = 3^x

3^2(z-2x)=3^x

so x is

2(z-2z)=x

z is

2z-4z=x

-2z=x

z=x/-2

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#9

(Original post by

wot level is this? i dont know what the first one is

The second one is (i think)

9^(z-2x) = 3^x

3^2(z-2x)=3^x

so x is

2(z-2z)=x

**Tazzie**)wot level is this? i dont know what the first one is

The second one is (i think)

9^(z-2x) = 3^x

3^2(z-2x)=3^x

so x is

2(z-2z)=x

**x**) = x

0

the function f(x) is defined by

f(x) = (2 + sin2x)/(2 + cosx)

Ignoring terms in x^3 and higher powers of x, obtain a quadratic approximation to the function f(x) for small values of x.

i tried to work out and got to an answer but i dont think its good becuase it was very easy, can somebody work it out to check pls?

i thank u in advance Kris

f(x) = (2 + sin2x)/(2 + cosx)

Ignoring terms in x^3 and higher powers of x, obtain a quadratic approximation to the function f(x) for small values of x.

i tried to work out and got to an answer but i dont think its good becuase it was very easy, can somebody work it out to check pls?

i thank u in advance Kris

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#11

I get f(x) = 2/3 + (2/3).x - x²/9

Looking at the original expression, you can see that f(x) -> 2/3 as x-> 0, which tends to confirm the expression I got above.

Looking at the original expression, you can see that f(x) -> 2/3 as x-> 0, which tends to confirm the expression I got above.

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#13

(Original post by

I get f(x) = 2/3 + (2/3).x - x²/9

Looking at the original expression, you can see that f(x) -> 2/3 as x-> 0, which tends to confirm the expression I got above.

**Fermat**)I get f(x) = 2/3 + (2/3).x - x²/9

Looking at the original expression, you can see that f(x) -> 2/3 as x-> 0, which tends to confirm the expression I got above.

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#14

(Original post by

Agree with Fermat, except isn't the coefficient of x^2 positive?

**shiny**)Agree with Fermat, except isn't the coefficient of x^2 positive?

f(x) = (2 + sin2x)/(2 + cosx)

sinx = x - x³/3! + x^5/5! - ...

cosx = 1 - x²/2! + x^4/4! - ...

excluding powers of x³ and higher, then substitute for these series for sinx and cosx into the expression for f(x).

f(x) = (2+2x)/(2+1-x²/2)

f(x) = (4+4x)/(6-x²)

f(x) = (1/6).(4+4x)/(1-(x/root(6))²)

Now, using the binomial series we can expand the expression (1 + x)^n to get,

(1-(x/root(6))²)^-1 = 1 + (-1).(-x/root(6))² + ... ) (this is where I had the sign wrong!

(1-(x/root(6))²)^-1 = 1 + (-1).(-x/root(6))²

(1-(x/root(6))²)^-1 = 1 + (-1).(-x²/6)

(1-(x/root(6))²)^-1 = 1 + (x²/6)

Now,

f(x) = (1/6).(4 + 4x).(1 + x²/6)

f(x) = (1/6).(4 + (4/6)x² + 4x + (4/6)x³), now ignore the term in x³

f(x) = (1/6).(4 + 4x + (2/3)x²)

f(x) = 2/3 + (2/3).x + (1/9).x²

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