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Old 04-02-2010   #1
magmagod
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Default Isolate to solve the equation

So the question is basically solve each equation for 0<x<2pi...i just have trouble with isolating the equations for the following:

square root 2 * sinx = 2sinx cosx

2cosx - 3/cosx + 2 * square root2 = 0

for the first one i got this far:
square root 2 sinx = 2sinx (double angle)
squared both sides and ended with 2 sinx = 4 sin^2x
divided by 2 sinx to both sides and got 1 = 2 sinx
then i isolated for sinx ... sinx = 1/2
is it right so far? then i just found the related acute angle and the values in Q 1 and 3

I appreciate your help so far i will try and help you with your problems..

Last edited by magmagod; 04-04-2010 at 12:01 PM..
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Old 04-03-2010   #2
MAS1

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Quote:
Originally Posted by magmagod View Post
So the question is basically solve each equation for 0<x<2pi...i just have trouble with isolating the equations for the following:

- square root 2 sinx = 2sinx cos x

- 2cosx - 3/cosx + 2 square root2 = 0

for the first one i got this far:
square root 2 sinx = 2sinx (double angle)
squared both sides and ended with 2 sinx = 4 sin^2x
divided by 2 sinx to both sides and got 1 = 2 sinx
then i isolated for sinx ... sinx = 1/2
is it right so far? then i just found the related acute angle and the values in Q 1 and 3

I appreciate your help so far i will try and help you with your problems..
Wow, these are tough problems the way you have presented them. Maybe you have left out some parentheses or something.

For the 1st one, I am assuming that first dash is NOT a negative sign, therefore:
sinx*sqrt(2) = 2sinxcosx
sqrt(2) = 2cosx
sqrt(2)/2 = cosx
Taking the inverse cos of both sides gives
x = pi/4 and 7*pi/4

If the first dash is a negative sign, then x = 3*pi/4 and 5*pi/4.

Your solution is incorrect because 2sinxcosx = sin2x, not 2sinx.
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Old 04-04-2010   #3
MAS1

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Originally Posted by MAS1 View Post
Wow, these are tough problems the way you have presented them. Maybe you have left out some parentheses or something.

For the 1st one, I am assuming that first dash is NOT a negative sign, therefore:
sinx*sqrt(2) = 2sinxcosx
sqrt(2) = 2cosx
sqrt(2)/2 = cosx
Taking the inverse cos of both sides gives
x = pi/4 and 7*pi/4

If the first dash is a negative sign, then x = 3*pi/4 and 5*pi/4.

Your solution is incorrect because 2sinxcosx = sin2x, not 2sinx.
I see you edited your equations.

For the second one:
2cosx - 3/cosx + 2*sqrt(2) = 0
cosx*[2cosx - 3/cosx + 2*sqrt(2)] = cosx*0
2(cosx)(cosx) + 2sqrt(2)*cosx - 3 = 0
Using quadratic formula to solve for cosx gives:
cosx = sqrt(2)/2 and cosx = -3*sqrt(2)/2
For the 1st solution x = pi/4 and 7*pi/4
For the 2nd solution x = arccos(-3*sqrt(2)/2)

Sorry, I don't have a calculator with trig functions on it for the second solution.

MAS1
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Old 04-05-2010   #4
jmw106462

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sinx*sqrt(2 = 2sinxcosx
sqrt(2 = 2cosx
sqrt(2/2 = cosx

x = pi/4 and 7*pi/4


This is what the answer is very hard like mas1 said. Mostly the way you presented them, didn't queit get the problem at first. Hope you do well, good luck!
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