Multiply both sides by sinx
1 + cosx = sinx
sinx - cosx = 1
(1/√2)sinx - (1/√2)cosx = (1/√2)
sinx cos(π/4) - sin(π/4) cosx = (1/√2)
sin(x - π/4) = (1/√2)
x - π/4 = π/4, 3π/4, 9π/4, 11π/4 etc
hence x = π/2, 5π/2 etc
Although the previous line also leads to values such as π, 3π these are not solutions as for those values sinx = 0 so that cscx and cotx are undefined.
Alternatively, express in terms of t, where t = tan(x/2)
(1+t²)/2t + (1-t²)/2t = 1
(1+t²) + (1-t²) = 2t
2 = 2t
t = 1
Hence x/2 = π/4, 5π/4, 9π/4, 13π/4 etc
x = π/2, 5π/2, 9π/2 etc
ONE MORE possibility is to write
cscx = 1 - cotx
then square both sides,
then replace csc²x by 1+cot²x giving an equation in cotx.
However solving this will, I guess, like the first method, introduce extraneous roots so you'll have to check and see which ones are valid. (But we know already they are π/2, 5π/2, 9π/2 etc)
1 + cosx = sinx
sinx - cosx = 1
(1/√2)sinx - (1/√2)cosx = (1/√2)
sinx cos(π/4) - sin(π/4) cosx = (1/√2)
sin(x - π/4) = (1/√2)
x - π/4 = π/4, 3π/4, 9π/4, 11π/4 etc
hence x = π/2, 5π/2 etc
Although the previous line also leads to values such as π, 3π these are not solutions as for those values sinx = 0 so that cscx and cotx are undefined.
Alternatively, express in terms of t, where t = tan(x/2)
(1+t²)/2t + (1-t²)/2t = 1
(1+t²) + (1-t²) = 2t
2 = 2t
t = 1
Hence x/2 = π/4, 5π/4, 9π/4, 13π/4 etc
x = π/2, 5π/2, 9π/2 etc
ONE MORE possibility is to write
cscx = 1 - cotx
then square both sides,
then replace csc²x by 1+cot²x giving an equation in cotx.
However solving this will, I guess, like the first method, introduce extraneous roots so you'll have to check and see which ones are valid. (But we know already they are π/2, 5π/2, 9π/2 etc)