secx tanx=1

1/cosx sinx/cosx=1

1 sinx/cosx=1

1 sinx=cosx

sinx=cosx-1

sinx=(-sinx)

2sinx=0

sinx=0

x=0, pi, 2pi

secx tanx = 1 find all solutions algebraically in the interval [0,2pi).

HERE

1/cosx sinx/cosx = 1

1 sinx = cosx = sqrt(1 sinx^2)

Squaring both sides

1 2sinx sinx^2 = 1 sinx^2

2(sinx)^2 2sinx = 0

2sinx(sinx 1) = 0

Here, Either

sinx = 0, For which x = 0, pi, qand 2pi.

OR

sinx 1 = 0, For which, x = -3pi/2

Thus

x = 0, pi, -3pi/2, and 2pi. Answer

sec(x) = 1/cos(x)

tan(x) = sin(x)/cos(x)

sec(x) tan(x) = 1

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

*Note here that by doing this step, cos(x) = 0 isnt allowed any more*

=> 1 sin(x) = cos(x)

=> cos(x) cos(x pi/2) = 1

Which of course only happens when either:

cos(x) = 0 and cos(x pi/2) = 1, but cos(x) = 0 isnt allowed

cos(x) = 1 and cos(x pi/2) = 0

So cos(x) = 1 => x = 0 is the only solution (2pi is outside the given interval)

Its mildly attractive youre doing this sort of math, not going to lie.

This post is last updated on hrtanswers.com at Date : 1st of September – 2022