Alan

Square both sides:

4cos²x = sin²x + 2sinx + 1

Replace cos² by 1 - sin²

4(1 - sin²x) = sin²x + 2sinx + 1

Collect terms

5sin²x + 2sinx - 3 = 0

This can be written as

(sinx + 5/5)(sinx - 3/5) = 0  or (sinx + 1)(sinx - 3/5) = 0

So sinx = -1 and sinx = 3/5

This means x = asin(-1) = 3pi/2 (=270°)

and x = asin(3/5) ≈ 36.87°

Adding and subtracting multiples of 2pi (360°) to these also satisfies the original equation.

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Use the fact that $$\int{x^n}dx=\frac{x^{n+1}}{n+1}$$ as long as n ≠ -1.

So $$\int{(5x^2-5x+9)dx}=\frac{5x^3}{3}-\frac{5x^2}{2}+9x+k$$ where k is an arbitrary constant of integration.

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Deleted

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I think you need to know the height of the satellite here!

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Melody, your answer is certainly simpler!

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k = (1/L)*ΔL/ΔT  

ΔT = (1/L)*ΔL/k

Brass ΔT = (1/2.583)*(D-2.583)/(19*10^-6)

Steel ΔT = (1/2.585)*(D-2.585)/(11*10^-6)

By dividing the Brass by the Steel we get:

1 = (2.585/2.583)*(D-2.583)/(D-2.585)*11/19

This gives D ≈ 2.588 cm

Plug this back in to one of the equations above to get ΔT ≈ 96.89degC

32°C + 96.89°C = 128.89°C ≈ 129°C

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The coldest point will be at the minimum (the lowest point) of the surface defined by the function of x and y.  The gradient of this surface in the x-direction is zero there, as is the gradient in the y-direction. These conditions can be used to find the values of x and y, which, when plugged back into the expression fgor temperature will give you the minimum value.

tan(2θ) = 2*tan(θ)/(1- tan²(θ))

Taking the limit as θ→pi/2 we have 2*tan(pi/2)/(1- tan²(pi/2)) = tan(pi) = 0.

(Geno, I think you got a sign wrong: should be 1 + tan² = sec² in your fourth line)

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Young's modulus (E) is Stress/Strain

Stress = force/unit area = mass*g/(pi*radius²)

Strain = change in length/original length = δL/L

 Hence:

δL = L*mass*g/(pi*radius²*E)

$${\mathtt{deltaL}} = {\frac{{\mathtt{0.38}}{\mathtt{\,\times\,}}{\mathtt{147}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{0.021}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{9}}}\right)}} \Rightarrow {\mathtt{deltaL}} = {\mathtt{0.000\: \!043\: \!903\: \!185\: \!783\: \!3}}$$

Note that I've turned length measurements into metres.

δL ≈ 4.4*10^-5m  = 0.044 mm

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