PreCal
posted by Lilly .
How do you find the number of solutions(not all the answers) for each of the following equations
1) 2sec^2 4x  3sec4x =5
2) 2tan^2 12x +3tan12x 7 =0
3) 2sin^2 5x + 3sin5x +1 =0

PreCal 
Steve
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.
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