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Q: What is the amplitude of the function y 3 sin 3x?
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What is the amplitude of the function y -3 sin 3x?

amplitude of the function y =-3 sin 3x


Parentheses before -1 and after 3x and before 3 and after 3x What is the answer to -2 cos2 3x-3 sin 3x equals 0?

I think you mean to solve:(-2 cos23x) - (3 sin 3x) = 0cos2 x + sin2 x = 1⇒ cos2x = 1 - sin2 x⇒ -2 cos2 3x - 3 sin 3x = -2(1 - sin2 3x) - 3 sin 3x = 0⇒ 2 sin2 3x - 3 sin 3x - 2 = 0⇒ (2 sin 3x + 1)(sin 3x - 2) = 0⇒ sin 3x = -1/2 or 2sin 3x = 2 is impossible as the range of sine is -1 ≤ sine ≤ 1Thus:sin 3x = -1/2⇒ 3x = 2nπ - π/6 or 2nπ - 5π/6⇒ x = 2/3nπ -π/18 or 2/3nπ -5π/18


How do you use the Euler's formula to obtain the sin3X in terms of cosX and sinX?

Using Euler's Formula, you use (cos(x) + i sin(x))^n = cos (nx) + i sin(nx) Now you let n=3 (cos(x) + i sin (x))3 = cos(3x) + i sin (3x) (cos(x))3 + 3(cos(x))2 * i sin(x) + 3cos(x) * i2 (sin(x))3 = cos(3x)+ i sin(3x) (cos(x))3 + i(3sin(x)(cos (x))2) - 3cos(x)(sin(x)2) - i(sin(x))3 = cos (3x) + i sin(3x) Now only use the terms with i in them to figure out what sin(3x) is... 3sin(x)(cos(x))2 - (sin(x))3 = sin(3x) Hope this helps! :D


Find the derivatives of y equals 2 sin 3x and show the solution?

y = 2 sin 3x y' = 2(sin 3x)'(3x)' y' = 2(cos 3x)(3) y' = 6 cos 3x


How would you evaluate the indefinite integral -2xcos3xdx?

Integrate by parts: ∫ uv dx = u ∫ v dx - ∫ (u' ∫ v dx) dx Let u = -2x Let v = cos 3x → u' = d/dx -2x = -2 → ∫ -2x cos 3x dx = -2x ∫ cos 3x dx - ∫ (-2 ∫ cos 3x dx) dx = -2x/3 sin 3x - ∫ -2/3 sin 3x dx = -2x/3 sin 3x - 2/9 cos 3x + c


Find the derivatives of y equals 2 sin 3x?

y=2 sin(3x) dy/dx = 2 cos(3x) (3) dy/dx = 6 cos(3x)


What is the inverse function of y equals 3x-3?

6


What are function rules?

y=3x=3


How do you find the amplitude maximum minimum and period for y equals -1 plus 3sin4x?

y = -1 + 3 sin 4xLet's look at the equation of y = 3 sin 4x, which is of the form y = A sin Bx, wherethe amplitude = |A|, and the period = (2pi)/B.So that the amplitude of the graph of y = 3 sin 4x is |3| = 3, which tell us that the maximum value of y is 3 and the minimum value is -3, and the period is (2pi)/4 = pi/2, which tell us that each cycle is completed in pi/2 radians.The graph of y = -1 + 3 sin 4x has the same amplitude and period as y = 3 sin 4x, and translates the graph of y = 3 sin 4x one unit down, so that the maximum value of y becomes 2 and the minimum value becomes -4.


Integration cos3 x?

Take f(x) = cos(3x) ∫ f(x) dx = ∫ cos(3x) dx Take u=3x → du = 3dx = ∫ 1/3*cos(u) du = 1/3*∫ cos(u) du = 1/3*sin(u) + C, C ∈ ℝ = 1/3*sin(3x) + C


How do you write this equation 3x equals 2y plus 3 in function form?

3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3


How do you write 3X equals 2Y plus 3 in function form?

3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3