# trig check!

• Apr 13th 2007, 10:05 AM
bay
trig 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!:)
• Apr 13th 2007, 02:20 PM
Jhevon
Quote:

Originally Posted by bay
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!:)

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.

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
• Apr 13th 2007, 03:10 PM
Plato
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.
• Apr 13th 2007, 03:13 PM
Jhevon
Quote:

Originally Posted by Plato
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.

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