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

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


tan%282x%29%2Bsec%282x%29+=+7
sin%282x%29%2Fcos%282x%29%2B1%2Fcos%282x%29+=+7
%28sin%282x%29%2B1%29%2Fcos%282x%29+=+7
sin%282x%29%2B1+=+7cos%282x%29
7cos%282x%29-sin%282x%29+=+1

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

49%28cos%282x%29%29%5E2-14cos%282x%29sin%282x%29%2B%28sin%282x%29%29%5E2+=+1
48%28cos%282x%29%29%5E2-14cos%282x%29sin%282x%29+=+0
cos%282x%29%2848cos%282x%29-14sin%282x%29%29+=+0
cos%282x%29=0 or 48cos%282x%29-14sin%282x%29=0

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

48cos%282x%29-14sin%282x%29=0
48cos%282x%29=14sin%282x%29
48%2F14+=+sin%282x%29%2Fcos%282x%29+=+tan%282x%29

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