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Old 04-04-2010   #3

Join Date: Dec 2008
Posts: 249

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 is offline