Question

Find all solutions in the interval [0, 2pi).
tan x + sec x = 1

Answer 1

MY WORK:

(sin x/ cos x) + (1/ cos x) = 1
(sin x + 1)/ cos x = 1

Then, I'm stuck. :(

Answer 2

subtract sec(x) from both sides first

tan(x) = 1 - sec(x)

then square both sides

tan^2(x) = (1-secx)^2 = 1 - 2secx + sec^x

then use the trigonometric identity tan^(x) = sec^2(x)-1 to change the left side

sec^2(x) - 1 = 1 - 2secx + sec^x

can you finish from here?

Answer 3

ah, a bit of a typo

last expression should be

sec^2(x) - 1 = 1 - 2secx + sec^2(x)

Answer 4

So what I was doing is wrong?

Answer 5

well, not "wrong" but it would be difficult to get an answer by converting tan(x) and sec(x) into sin(x) and cos(x)

Answer 6

there might be a way to do it your way, but I can't think of an easy way to go from there

Answer 7

Wait, so will it be equal to 0?

Answer 8

sec^2(x) - 1 = 1 - 2secx + sec^2(x)

cancel out sec^2(x)

-1 = 1 - 2sec(x)
2sec(x) = 2
sec(x) = 1

then solve for x

Answer 9

Then it is 0. :D

Answer 10

yea