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

Solution:

Given, tan x + sec x = 1

We have to find all the solutions in the interval [0, 2π)

By using trigonometric identity,

sec²x - tan²x = 1

So, sec²x = 1 + tan²x

sec x = √(1 + tan²x)

Now, tan x + √1 + tan²x = 1

On rearranging,

√1 + tan²x = 1 - tan x

Squaring both sides,

1 + tan²x = (1 - tan x)²

1 + tan²x = 1 + tan²x - 2 tan x

By grouping,

1 - 1 + tan²x - tan²x = -2tan x

-2 tan x = 0

tan x = 0

Taking inverse,

x = tan^-1(0)

x = 0, π

sec π = -1, so π is an extraneous root.

Therefore, the solution is 0.

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

Summary:

All solutions in the interval [0, 2π). tan x + sec x = 1 is 0.