Q: How do you solve sin n divided by n for n?

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Its just 8/n or 8 divided by n. You can't solve this problem because there is no other number to simplify or solve.

If sin2(theta) = 0, then theta is N pi, N being any integer

2n/3 = 4 2n = 3*4 2n = 12 n = 6 :)

sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x

Do sin(x), square it, and then multiply it by two.

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sinx=n/1 (1)sinx=n/1(1) sin(-n)x=n(-n) six=6

19

Its just 8/n or 8 divided by n. You can't solve this problem because there is no other number to simplify or solve.

sin(x)*sin2(x) = 1 so sin3(x) = 1 so that sin(x) = cuberoot(1) = 1 then x = pi/2 + n*pi where n is an integer.

5x = 80

If sin2(theta) = 0, then theta is N pi, N being any integer

19

1.5

2n/3 = 4 2n = 3*4 2n = 12 n = 6 :)

sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x

Do sin(x), square it, and then multiply it by two.

sin7x-sin6x+sin5x