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January 30, 2015

January 30, 2015

Posted by **Lilly** on Monday, November 18, 2013 at 8:59pm.

1) 2sec^2 4x - 3sec4x =5

2) 2tan^2 12x +3tan12x -7 =0

3) 2sin^2 5x + 3sin5x +1 =0

- Pre-Cal -
**Steve**, Tuesday, November 19, 2013 at 12:19amthese 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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