How do you solve the trig equation tan(x/2) + 1 = cosx?

tan(x/2) + 1 = cosx

tan(x/2) + 1 = 1- 2 sin ^2 (x/2)

sin (x/2) / cos (x/2) = - 2 sin ^2 (x/2)

sin (x/2) + 2 sin ^2 (x/2) cos (x/2) =0

sin (x/2) [ 1 + 2 sin (x/2) cos (x/2) ] =0

sin (x/2) =0 or 1 + 2 sin (x/2) cos (x/2) =0

x/2 = n pi

x = 2n pi

answer

1 + 2 sin (x/2) cos (x/2) =0

1 + sin x =0

sin x = -1

x = 3n pi/4

answer

1) Using half angle identity, cos(x) = {1 - tan²(x/2)}/{1 + tan²(x/2)}

2) So, let tan(x/2) = t, the given one transforms to:

t + 1 = (1-t²)/(1+t²)

==> 1 + t + t² + t^3 = 1 - t²

==> t(t² + 2t + 1) = 0

Solving this, either t = 0 or -1

3) So, tan(x/2) = 0 or -1

i) If tan(x/2) = 0, then x/2 = 0; ==> In general x/2 = nπ, where 'n' is an integer.

So, x = 2nπ, where 'n' is an integer

Note: If we consider x = nπ, for all 'n' to be odd integer, tan(x/2) = 0 and cos(x) = -1. This does not satisfy the given equation. Hence in general x = 2nπ only.

ii) If tan(x/2) = -1/2, then x/2 = -π/4; General solution, x/2 = nπ - π/4

==> x = 2nπ - π/2, where 'n' is an integer.

This satisfies both sides of the equation.

Thus from the above,

x = Either 2nπ or (4n-1)(π/2)

use identity :

cosx = (1-tan^2 (x/2))/(1+tan^2(x/2) and you get:

(1-tan^2 (x/2))/(1+tan^2(x/2) = tan(x/2) + 1

(1-tan^2 (x/2)) = (tan(x/2) + 1)(1+tan^2(x/2)

1-tan^2 (x/2) = tan(x/2) +tan^3(x/2)+1 + tan^2 (x/2)

- tan^2 (x/2) = tan(x/2) + tan^3(x/2) + tan^2 (x/2)

tan(x/2) + tan^3(x/2) + 2tan^2 (x/2) = 0

tan(x/2)[1 + tan^2(x/2) + 2tan (x/2)] = 0

tan(x/2)[1 +tan (x/2)]^2 = 0

you must solve

tan(x/2)=0

tan (x/2)= - 1