Question

Find all solutions in the interval [0, 2π).

tan x + sec x = 1

Answer 1

help me!

Answer 2

As I am not used to the sec function, I would begin with:\[\tan x+\sec x=1 \Leftrightarrow \frac{ \sin x }{ \cos x }+\frac{ 1 }{ \cos x }=1\]Then write it as one fraction.
Give it a try!

Answer 3

@ZeHanz is it pi/4?

Answer 4

No, I have found several solutions on [0, 2π), but none of them is pi/4.

Answer 5

@ZeHanz 5pi/4?

Answer 6

Sorry, no...

Answer 7

ohh it's zero is it?

Answer 8

To give you a start, let me take a few steps:\[\frac{ \sin x + 1 }{ \cos x }=1\]This means:\[\sin x + 1 = \cos x\]Also:\[(\sin x + 1)^2=\cos^2x\]

Answer 9

I'm afraid there is some work to do before the solutions become visible...

Answer 10

He's just telling all the options! :D

Answer 11

or she*

Answer 12

@shubhamsrg ohh okay! so could 0 be one of them?

Answer 13

LAWL !

Answer 14

asking the wrong person! :|

Answer 15

@ZeHanz could 0 be of the the answers?

Answer 16

I think it is better to focus on finding the solutions.
I have taken a few steps, now it's up to you to go on with this.
It is much more rewarding fo find the solutions and then check what option to choose.

If you get stuck, just ask and I will explain.

Answer 17

@shubhamsrg: this is why I hate multiple choice questions.
Why can't makers of tests just ask to solve an equation?
I am a teacher myself, but here in Holland there are no multiple choice questions in maths.
The drawback: it takes time to correct tests students make.

Answer 18

I see your point. You are right :)

Answer 19

@ZeHanz but sometimes students learn by looking at the multiple choice options. Seeing the asnwer I got from solving the question on the listed option encourages me that Im solving the question the right way. When the teachers give all the questions in free response, even if i get 100 on the test, i still question my answers because not all the teachers go over every problems on the test. just sharing my opinion as a student..

Answer 20

trial and error ^^

Answer 21

With this kind of equations, even giving away all the solutions wouldn't make the process of solving easier. You still have to solve the equation.

Sometimes you can see a solution without calculation. That's ok, but because there may be other solutions, there is still a calculation needed. The outcome of this calculation will give all solutions, including the one(s) you saw right away.


BTW to get further with (sin x+1)²=cos²x, you need to expand the left hand side and replace cos²x with 1-sin²x. Please try it!