SOLUTION: Hi. I need help with the following problem. "Find the solutions of the equation in interval [0,2pi)." the problem is secx+tanx=1. The only thing I did was subtract the one, but now

Question 36014: Hi. I need help with the following problem. "Find the solutions of the equation in interval [0,2pi)." the problem is secx+tanx=1. The only thing I did was subtract the one, but now i'm unsure what to do. Please help!
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Hi. I need help with the following problem. "Find the solutions of the equation in interval [0,2pi)." the problem is
secx+tanx=1...............................I
WE KNOW THAT SEC^2(X)-TAN^2(X)=1={SEC(X)+TAN(X)}{SEC(X)-TAN(X)}
HENCE FROM EQN.I ,WE GET THAT
SEC(X)-TAN(X)=1................................II
EQN.I+EQN.II....GIVES
2SEC(X)=2
SEC(X)=1
COS(X)=1
X=0......ONLY AS THE NEXT SOLUTION 2PI IS EXCLUDED FROM THE RANGE
The only thing I did was subtract the one, but now i'm unsure what to do. Please help!
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