SOLUTION: solve tan 2x + sec 2x = 7 over the interval [0,2pi)

Question 1178771: solve tan 2x + sec 2x = 7 over the interval [0,2pi)
Found 2 solutions by Alan3354, greenestamps:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
solve tan 2x + sec 2x = 7 over the interval [0,2pi)
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Is is sec(2x)?
or sec^2(x)?
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

Square both sides and simplify using sin^2(A)+cos^2(A)=1

or

cos(2x)=0 is an extraneous solution; it does not satisfy the original equation (sec(2x) and tan(2x) are both undefined when cos(2x)=0). So

First solution:
2x = arctan(48/14) = arctan(24/7) = 1.287 radians (to a few decimal places);
x = 1.287/2 = 0.6435

The period of the function is the period of cosine(A), which is 2pi. In the prescribed interval for x, [0,2pi), 2x takes on the values 0 to 4pi. So there is a second solution on the prescribed interval.

Second solution:
2x = arctan(48/14)+2pi
x = 0.6435+pi = 3.7851

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