SOLUTION: solve tan 2x + sec 2x = 6 over the interval [0, 2pi)

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Question 1178825: solve tan 2x + sec 2x = 6 over the interval [0, 2pi)
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

solve tan+%282x%29+%2B+sec+%282x%29+=+6 over the interval [0, 2pi)
sin+%282x%29%2Fcos%282x%29+%2B+1%2Fcos%282x%29+=+6
%28sin+%282x%29%2B+1%29%2Fcos%282x%29+=+6+
%28sin+%282x%29%2B+1%29=+6cos%282x%29 ......square both sides
%28sin+%282x%29%2B+1%29%5E2=+%286cos%282x%29%29%5E2
sin%5E2+%282x%29%2B2sin+%282x%29%2B+1=+36cos%5E2%282x%29...............since cos%5E2%282x%29=1-sin%5E2%282x%29, we have
sin%5E2+%282x%29%2B2sin+%282x%29%2B+1=+36%281-sin%5E2%282x%29%29

sin%5E2+%282x%29%2B2sin+%282x%29%2B+1=+36-36sin%5E2%282x%29
sin%5E2+%282x%29%2B36sin%5E2%282x%29%2B2sin+%282x%29=36-1
37sin%5E2%282x%29%2B2sin+%282x%29-35=0.....use quadratic formula
sin+%282x%29=%28-2%2B-sqrt%282%5E2-4%2A37%28-35%29%29%29%2F%282%2A37%29
sin+%282x%29=%28-2%2B-sqrt%284%2B5180%29%29%2F74
sin+%282x%29=%28-2%2B-sqrt%285184%29%29%2F74
sin+%282x%29=%28-2%2B-72%29%2F74
solutions:
sin+%282x%29=%28-2%2B72%29%2F74=70%2F74=35%2F37
or
sin+%282x%29=%28-2-72%29%2F74=-74%2F74=-1
so,
2x=sin%5E-1+%28-1%29
2x=-pi%2F2+
x=-pi%2F4 (result in radians)
sin is negative in Q III or IV
x=215.54° (degrees)
or
2x=sin%5E-1+%2835%2F37%29
2x=1.24049897
x=1.24049897%2F2
x=0.620249485(result in radians)
in degrees:
x=35.54° (degrees)