Question

Find the general solution of the following equation:

cosθ + sinθ = 1.

Answer 1

cosθ + sinθ = 1

Dividing both sides by `sqrt((1)^2 + (1)^2) = sqrt(2)`, we get

`(1)/sqrt(2)cosθ + (1)/sqrt(2)sinθ = (1)/sqrt(2)`

`∴ cos  pi/(4) cosθ + sin  pi/(4) sinθ = cos  pi/(4)`

∴ `cos(θ - pi/4) = cos  pi/(4)`  ...(1)

The general solution of cosθ = cos α is θ = 2nπ ± α, n∈Z.

∴ The general solution of (1) is given by

`θ - pi/(4) = 2nπ ± pi/(4)`, n∈Z

Taking positive sign, we get

`θ - pi/(4) = 2nπ + pi/(4)`, n∈Z

∴ θ  = 2nπ + `π/2`, n∈Z

Taking negative sign, we get,

`θ - pi/(4) = 2nπ - pi/(4)`, n∈Z

∴ θ  = 2nπ, n∈Z

∴ The required general solution is θ  = `2nπ + pi/2`, n∈Z or θ = 2nπ, n∈Z.

Alternative Method:

cosθ + sinθ = 1

∴ sinθ = 1 – cosθ

∴ `2sin  θ/(2) cos  θ/(2) = 2sin^2  θ/(2)`

∴ `2sin  θ/(2) cos  θ/(2) - 2sin^2  θ/(2)` = 0

∴ `2sin  θ/(2)(cos  θ/2 - sin  θ/2)` = 0

∴ `2sin  θ/(2) = 0 or cos  θ/(2) - sin  θ/(2)` = 0

∴ `sin  θ/(2) = 0 or sin  θ/(2) = cos  θ/(2)`

∴ `sin  θ/(2) = 0 or tan  θ/(2) = 1       ...[∵ cos  θ/(2) ≠ 0]`

∴ `sin  θ/(2) = 0 or tan  θ/(2) = tan  pi/(4)    ...[∵ tan  pi/(4) = 1]`

The general solution of sin θ = 0 is θ = nπ, n∈Z and tan θ = tan α is θ  = nπ + α, n∈Z.

∴ The required general solution is `θ/(2)` = nπ, n∈Z or `θ/(2)` = nπ + `pi/(4)`, n∈Z

i.e. θ = 2nπ, n∈Z or θ = `2nπ + pi/(2)`, n∈Z.