if 1+sinx+sin^2x+..........to infinity=4+2undreroot3, 0<x<pi.find x?

Question

1)x=pi/3 or pi/6
2)x=pi/3 or 2pi/3

Answer 1

To elaborate on nayan's answer...


The left side is an infinite series whose ratio's absolute value is less than 1 (I know, sin x can equal 1 or -1, but if it did, then the series would not converge. The problem implies ocnvergence). Since the first term is 1 and the ratio is sin x, then the infinite sum on the left equals 1/(1 - sin x).

Now you have 1/(1 - sin x ) = 4 + 2*SQRT(3).

I would reciprocate both sides to get
1 - sin x = 1/(4 + 2*SQRT(3)).

While we're at it, I'd go ahead and rationalize the denominator on the right side. To do this, multiply the right side by
(4 - 2*SQRT(3)) / (4 - 2*SQRT(3)).

It's a messy FOIL problem, but if you pull it off, you'll get
1 - sin x = (4 - 2*SQRT(3)) / 4, or
1 - sin x = 1 - SQRT(3) / 2

Subtract 1 from both sides, and divide by -1 to get
sin x = SQRT(3) / 2.

I assume that by asking a question of this caliber, you are already familiar with basic trig stuff. So it should be clear (or at least easily solvable) that x = pi/3 or 2pi/3 (remember: sine is positive in quadrant II).

Answer 2

2

Answer 3

1 + sinx ( 1 + sinx + sin^2x ...) = 4 + 2 rt 3
1 + sinx (4 + 2 rt 3) = 4 + 2 rt 3
sinx = (3 + 2 rt 3) / (4 + 2 rt 3)
= 2 rt 3 / 4
= rt 3 / 2

Ans : (2) x = pi / 3 or x = 2 pi / 3

Answer 4

1. * x=pi/3
. . pi / 3 = 1.047198.
. * x-pi/6
. . pi / 6 = 0.523599.

2. *2pi/3
. . pi / 3 = 1.047198.

if 1+sinx+sin^2x+..........to infinity=4+2undreroot3, 0 0 * pi = 0

Answer 5

That's too easy. It's the 6th character in the equation starting from the left.

Answer 6

yes, it is bad to sin

(how in the world could u trust people on Yahoo answers to do your homework?)

Answer 7

I thought this site was for posing questions.

What IS this?!!!

Answer 8

my cat's breath smells like cat food.