SOLUTION: Solve tanx+1=secx and 2tanxsecx-tanx=0

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Question 1078827: Solve tanx+1=secx and 2tanxsecx-tanx=0
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Is it 1 problem? Or 2?

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Solve
tan%28x%29%2B1=sec%28x%29

We square both sides

tan%5E2%28x%29%2B2tan%28x%29%2B1=sec%5E2%28x%29

We use the Pythagorean trig identity: 
1%2Btan%5E2%28theta%29=sec%5E2%28theta%29 to write
the right side:

tan%5E2%28x%29%2B2tan%28x%29%2B1=1%2Btan%5E2%28x%29

2tan%28x%29=0

tan%28x%29=0

You finish. That gives you two values for x in [0,2p).  
But since I squared both sides, you must check for 
extraneous solutions. 

--------------------------

2tan%28x%29sec%28x%29-tan%28x%29=0

Factor out common factor of tan(x)

tan%28x%29%282sec%28x%29-1%5E%22%22%29=0



You finish by solving tan(x)=0 for x.
But remember that sec(x) always has
absolute value greater than or equal to 1.

Edwin