Solve from [0,2pi) 2cos(x/2)-sin(x)=0

Answer 1

Hi James B.,

If you make a graph of the equation you will see that the only zero is at π.

From 0 to 2π:

x = 0, y = 2;

x = π/4, y = 1.14;

x = π/2, y = .41;

x = 3π/4, y = .06;

x = π, y = 0;

x = 5π/4, y = -.06

x = 3π/2, y = -.41

x = 7π/4, y = -1.14

x = 2π, y = -2

Plug π into the equation:

2cos(π/2) - sin(π) = 0,

0 - 0 = 0, true.

I hope this helps, Joe.

EDIT-------------------------------------------------------------------------------------------------------------------------------

Hi James B.,

x = 0; 2cos(0) - sin(0) = 2*1 - 0 = 2, graph (0, 2)

x = π/2; 2cos(π/4) - sin(π/2) = √2 - 1, graph (1.57, .41)

x = π; 2cos(π/2) - sin(π) = 0 - 0 = 0, (answer) graph (3.14, 0)

x = 3π/2; 2cos(3π/4) - sin(3π/2) = -√2 - -1 = -√2 + 1, graph (4.71, -.41)

x = 2π; 2cos(π) - sin(2π) = -2 - 0 = -2, graph (3.28, -2)

No calculator and all points can be plotted on a graph. The graph is sort of a backwards S, (Ƨ).

These answers are the same as above except all the π/4's are omitted because π/8's (for cos[π/2]) make things more complicated.

Any more questions fell free to ask me. I want you to have a clear understanding.

Thanks, Joe.