first off I want to say thanks for checking my math

http://www.mathhelpforum.com/math-he...onometrey.html

I was wondering if you could also check this! thanks once again!:)

Printable View

- April 13th 2007, 10:05 AMbaytrig check!
first off I want to say thanks for checking my math

http://www.mathhelpforum.com/math-he...onometrey.html

I was wondering if you could also check this! thanks once again!:) - April 13th 2007, 02:20 PMJhevon
Your work is in orange, My words are in blue:

Some of your work might not show up, since the symbols were pictures and not text.

Determine the general solution to sin x/2 = 3/ 2. Use radian measure in your solution.

The solution to the equation sin x/2 = 3/ 2 is given by intersection points, of both graphs. Points of intersection occur at x=120° and 239° when domain is 0°__<__x< 360°. Since the period is 360°, the general solutions are:

N= any integer x= 120° + n (360°) and x=239°+ n (360°)

Which then would together form the equation x= (360°) n__+__120°

I would have done this using the half angle formula(did you learn that?):

sin(x/2) = sqrt[(1 - cosx)/2]

=> sqrt[(1 - cosx)/2] = sqrt(3)/2

=> (1 - cosx)/2 = ¾

=> 1 - cosx = 3/2

=> cosx = 1 – 3/2 = -1/2

=> x = cos^-1(-1/2)

=> General Solution: x = 2pi/3 + 2kpi, x = 4pi/3 + 2kpi, for k some integer.

Since I squared the equation, let's see if any solution is extraneous:

Check 2pi/3:

=> sin(pi/3) = sqrt(3)/2 ........Check!

Check 4pi/3:

=> sin(2pi/3) = sqrt(3)/2 ......Check, both answers check out.

So your answers should be in radians, as mine are. Also, it's 240 degrees not 239.

Doing it your way you'd have to be familiar with the sine graph to know that sin(theta) = sqrt(3)/2 when theta = pi/3 and theta = 2pi/3

so x/2 = pi/3 => x = 2pi/3

or x/2 = 2pi/3 => x = 4pi/3

and we add 2kpi to each of these for the general solution.

2) Solve for x, where 0<x<2.

a. 2 tan x + 12 = 0

2 tan x + 12 = 0

2 tan x/2 + 12/2 = 0

Tan x= - 3

X= 2/3, 5/ 3

I wouldn't have done this much different. Well, not really, I would never have switched to tan(x/2), there's no need for a half angle.

2tanx + sqrt(12) = 0

=> 2tanx = - sqrt(12)

=> tanx = -sqrt(12)/2

=> tanx = -sqrt(4)sqrt(3)/2

=> tanx = - sqrt(3)

=> x = tan^-1(-sqrt(3))

=> x = 2pi/3, 5pi/3

b. 2x + cos x=1

2cos^2x + cos x=1

2cos^2x + cos x-1=0

(2cosx+1)(cosx-1)=0

2cosx+1=0

cosx=-1/2

cosx -1=0

cosx=1

x= /3,, 5/ 3,

You factorized incorrectly:

2cos^2x + cosx – 1 = 0

=> (2cosx – 1)(cosx + 1) = 0

=> 2cosx – 1 = 0

=> cosx = ½

=> x = pi/3 + 2kpi, x = 5pi/3 + 2kpi, for k some integer .........general solution

or

cosx + 1 = 0

=> cosx = -1

=> x = pi + 2kpi, for k any integer .........general solution

So x= pi + 2kpi, pi/3 + 2kpi, x = 5pi/3 + 2kpi, for k any integer .........general solutions

Write the general solutions to the equation 2.a. and 2.b.

2a) 2/3+ n,5/ 3+ n are the two general solutions

2B) 5/ 3+2n, n= any integer

2a) is fine, 2b) is incorrect, see the problem above - April 13th 2007, 03:10 PMPlato
Come on there is no solution to any equation such as sin(x/2)=(3/2)>1.

The sine function has no value greater than 1. - April 13th 2007, 03:13 PMJhevon
You are correct Plato, but it is not 3/2 it is actually sqrt(3)/2. it's just that in his word file, his squareroot signs are pics, not texts, so when i pasted them here, they didn't show up. see the word file he attached for the full questions, a lot of symbols were left out