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04022010  #1 
Join Date: Mar 2010
Posts: 9

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; 04042010 at 12:01 PM.. 
04032010  #2  
Join Date: Dec 2008
Posts: 249

Quote:
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. 

04042010  #3  
Join Date: Dec 2008
Posts: 249

Quote:
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 

04052010  #4 
Join Date: Feb 2010
Posts: 388

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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