What Are the Bounds of g(t) and Solutions for Complex Equations in Trigonometry?

[Danielll]

Hi I've been working on my assignment for ages and am down to the last two questions (finally) which I've been working on for ages and can't work out.

1. For g(t) = 2sint+cos2t over 0 <(equal to or greater than) x < (equal to or greater than) 4pi
show that -3 < g(t) < 3/2 (< greater than or equal to) for all t.

2.a) For x and y real, solve the equation iy/(1+ix) - (3y + 5i)/(3x + y) = 0
b) Z = (1 + iw)/(1 + iw - w^2). Assume w > (greater than or equal to). Find and then sketh |Z| and arg Z as functions of w.

I really need help with these asap. Thanks in advance.

 

[dextercioby]

Do you know calculus?It might help for the first problem...

As for 2a),by equating to 0 the # in the LHS,u'll get a system for "x" and "y".

Daniel.

 

[thepatientmental]

Hi there,

I understand your frustration and the urgency in needing help with your assignment. I am happy to assist you with these questions.

1. To show that -3 < g(t) < 3/2 for all t, we can use the fact that the maximum value of sin t and cos 2t is 1. Therefore, the maximum value of g(t) is 2+1 = 3. Similarly, the minimum value of sin t and cos 2t is -1, so the minimum value of g(t) is 2-1 = 1. This means that -3 < g(t) < 3 for all t. To show that g(t) is also less than or equal to 3/2, we can use the fact that the maximum value of sin t and the minimum value of cos 2t is 1/2. This means that the maximum value of g(t) is 2+1/2 = 5/2, which is less than 3. Therefore, we can conclude that -3 < g(t) < 3/2 for all t.

2.a) To solve the equation iy/(1+ix) - (3y + 5i)/(3x + y) = 0, we can rearrange the equation to get: iy(3x+y) - (3y+5i)(1+ix) = 0. Expanding the brackets, we get: 3ixy + iy^2 - 3y - 5i - 3xy - 5ix = 0. Simplifying, we get: (3x-1)y + (x+1)y^2 - 5i = 0. This is a quadratic equation in terms of y, so we can use the quadratic formula to solve for y. Plugging in the values, we get: y = (-3x+1 +/- sqrt((3x-1)^2 + 4(x+1)5i))/(2(x+1)). This gives us two solutions for y, which we can substitute back into the original equation to solve for x.

b) To find |Z| and arg Z as functions of w, we first need to simplify Z. We can use the fact that (a+b)(a-b) = a^2 - b^2 to simplify the denominator. This gives us: Z = (1 +