2pi

1. Solve each equation on the interval 0 ≤ θ< 2pi Secx=tanx=cotx

I'm having trouble with this
2. Solve on interval [0, 2pi): (cos x+1) (2cos^2 x-3 cos x-2) =0?

Solve on interval [0, 2pi): (cos x+1) (2cos^2 x-3 cos x-2) =0? I think B. x=pi, x=2pi/3 and x=4pi/3 is the answer is this correct. Thank you!
3. Solve on the interval [0, 2pi): 3sec(x-1)=2

Solve on the interval [0, 2pi): 3sec(x-1)=2 I think that this is A. 0 is this correct? Thank you!
4. Solve on the interval [0, 2pi): 1 + cos θ = 1/2?

Solve on the interval [0, 2pi): 1 + cos θ = 1/2? Let theta = x. So 1 + cos x = 1/2. Therefore, cos x = -1/2. We know that cos x = 1/2 at pi/3. However, cos x = -1/2 occurs in quadrants 2 and 3. Let pi/3 be the reference angle. So pi+pi/3 = 4pi/3 and pi - pi/3 = 2pi/3. So B is your answer...
5. Solve on the interval [0, 2pi): 4 csc x + 6= - 2

Solve on the interval [0, 2pi): 4 csc x + 6= - 2 Is this right below? 4csc(x) + 6 = -2 4csc(x) = -8 csc(x) = -2 sin(x) = -1/2 x = 7pi/6 , 11pi/6 The answer is D. then? Thank you!
6. Find all values of θ in the interval [0, 2pi]

Find all values of θ in the interval [0, 2pi], given: cot θ = -1/√3 cosθ= 1/2 I don't really need the answer, just need to know how to solve this type of problem :( I greatly appreciate your answer :) Thank you.
7. Need to find all values of X in the interval [0, 2pi]

Find all values of X in the interval [0,2pi] sin 2x = 3 cos x
8. C[0,2pi] separable?

So I know how to prove that the space of all continuous functions in [0,1] is separable. But I was thinking, the space of real valued continuous functions which are periodic (with period 2pi) should also be separable. The prof didn't prove this (and I am not sure he will) but I am interested in...
9. Solve [0,2pi)

2 similar problems that are really challenging for me. 1) solve [0,2pi) sin 2x-sinx = 0 2) solve [0,2pi) tan^2x sinx= tan^2x Also how do i rewrite cos^3 x so that it does not contain power greater than 1? I don't think i can use double angle identity since it has power of 3.. not...
10. Find solutions equation in internal [0, 2pi]

cos 2x = cos x Well first thing I would have to do is convert this into a quadratic formula (meaning ax2+bx+c) Though, my teacher helped me get started, and told me to do this: 2cos2(x)-cos(x)-1 I was lost at how the cos 2(x) became 2cos2x Can anyone explain to me this? ---- Though, after...
11. Solve the Equation exactly over the interval [0, 2pi].

Solve the Equation exactly over the interval [0, 2pi] the problems i do not understand are: 2sin^2x + sin x - 1 = 0 2sinx cosx + cos(x) = 0 2sin^2x - sin x = 0 if anyone could help me i would appreciate it alot. thanks
12. Solve 2pi = 6 cos^-1 (3x)

Find the exact solution of the equation 2pi = 6 cos^-1 (3x) My answer: x= 1/6 Seems too simple to be correct
13. Solutions between (0,2pi]

I am just trying to get some help on how to get this equation right. I have it equal to 0. Anything will help thanks. sin5x-sinx=2cos3x sin4x-2cos3x=0 (I think this is correct, just need help moving on to the next step.
14. Measure of angle between 0 and 2pi

Hey, I don't even know where to begin to solve this problem. If sinx= -1/2 and cosx=radical3/2, find a possible measure for angle X between 0 and 2pi. Would this problem have anything to do with a 30/60/90 triangle? Thanks for any and all help
15. [SOLVED] Solutions From [0,2pi]

2sin(x) + csc(x) = 0 the answer in the book says that there is no solution, why?
16. having 2pi as the upper limit of an interval

i cant make this work e.g. \int_0^2\pi \theta d\theta the pi is mean to be the upper limit of the integral, not inside the integrand any helps? also i cant find out how to do the inequality sign
17. Find the value between 0 and 2pi

Find the values of A in the range between 0 and 2pi for which sinA+sin3A=cosA+cos3A I have been trying for two days and still can't get it yet. Any help is greatly appreciated. (Bow)
18. All solutions from [0, 2pi)

tan^2-secx = 1
19. find all solutions in the interval of [0, and 2pi)

A.) 3sec^2(x/2)-4sec(x/2)-4=0 B.) tan^2x = (-3/2)-secx
20. Find the extrema of f(x) = 2 sinx + cos 2x when 0 < x < 2pi

I am having trouble determining the extrema of the above function. I think the derivative is 2 cos x - 2 sin 2x At 2 cos x= 0 ; cos is pi/2 and/or 3 pi/2. At - 2 sin 2x = 0; - 2 sin 2x = pi and/or 2 pi. 2 cos x = pi, ---> cos x = pi/2 and/or 3 pi/4. - 2 sin 2x = -pi/2 ---> - pi/4 or - pi/2. If...