Please help with this math?

Question

If 0< theta <360*

x=theta

solve
sec^2(2x) + 2sec(2x) = 0

Answer 1

sec^2(2x) + sec(2x) = 0
=> sec(2x)[sec(2x) -- 1] = 0
when sec(2x) = sec (90deg) => x = 45deg, 315deg
when sec(2x) = 1 = sec(0), x = 0, 260 deg

Answer 2

Factor: [sec (2x)] [sec (2x) +2] = 0 Solve.

Setting the first factor to zero gives you an empty solution set.

Setting the 2nd factor to zero gives you cos (2x) = -1/2

So your related angle for 2x is 60 degrees, but since the cosine function is negative, you answer(s) --for the 2x arguement are in quadrant II and III.

x is either 60 or 120 degrees (making 2x respectively 120 or 240 -- which are quadrant II and III angles and have the related angle of 60)
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Addendum/correction 2x could also be 120+360 or 240+360
making 2 more possibilities for x: 240 and 300 in the universe you allowed for solution(s). These 4 solutions are each coterminal with either of the 60 degree angles in QII and III.

Answer 3

sec^2(2x) + 2sec(2x) = 0
sec(2x)(sec(2x) + 2) = 0
sec(2x) = 0, - 2
sec(2x) cannot equal 0, so
sec(2x) = - 2
cos(2x) = - 1/2
2x = 120°, 240°
x = 60°, 120°, 240°, 300°

Answer 4

since x lies between 0 and 2pi ( 0 to 360 degrees)

2x lies between 0 and 4pi (0 to 720 degrees)


sec^2(2x) + 2sec(2x) = 0

1/cos^2(2x) + 2/cos(2x) = 0

multiply with cos^2(2x)

1 + 2cos(2x) = 0

2cos(2x) = -1

cos(2x) = -1/2

2x = 2pi/3, 4pi/3, 8pi/3, 10pi/3

x = pi/3, 2pi/3, 4pi/3, 5pi/3