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Old 04-04-2010   #3
MAS1

 
Join Date: Dec 2008
Posts: 249
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Quote:
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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