Posted by Lilly on Monday, November 18, 2013 at 8:59pm.
these are all done the same way. You have to solve a quadratic in some trig function.
#1.
2sec^2 4x - 3sec 4x - 5 = 0
(2sec 4x - 5)(sec 4x + 1) = 0
So, sec 4x = 5/4 or sec 4x = -1/4
There are two angles for each solution in [0,360°) where this is true.
But that means there are 8 solutions, since if sec(4x) = 5/4, then sec(4x+360),sec(4x+720),sec(4x+1080) are also 5/4.
Or, reducing to just x, we have
sec 4x
sec 4(x+90)
sec 4(x+180)
sec 4(x+270)
So, since sec(.64350) = 5/4,
x = .16075 + k*pi/2, k=0..3
for the other solution, sec 4x = -1/4, that's bogus, since |sec x| >= 1.
Now do the others the same way.