Question

find the principal solution for sinx= 1​

Answer 1

$\huge \underline\mathfrak \red{answer} =$

sin x =1x=(2nπ+π/2) =(4n+1) π/2

more sum's:—

sin x = 0 x = nπ

cos x = 0 x = (nπ + π/2)

tan x = 0 x = nπ

cos x = 1 x = 2nπ

sin x = sin θ x = nπ + (-1)nθ, where θ ∈ [-π/2, π/2]

cos x = cos θ x = 2nπ ± θ, where θ ∈ (0, π]

tan x = tan θ x = nπ + θ, where θ ∈ (-π/2 , π/2]

sin2 x = sin2 θ x = nπ ± θ

cos2 x = cos2 θ x = nπ ± θ

tan2 x = tan2 θ x = nπ ± θ

MARK ME AS BRAINLISIT

✌✌✌

Answer 2

Answer:

\huge \underline\mathfrak \red{answer} =

answer

=

sin x =1x=(2nπ+π/2) =(4n+1) π/2

more sum's:—

sin x = 0 x = nπ

cos x = 0 x = (nπ + π/2)

tan x = 0 x = nπ

cos x = 1 x = 2nπ

sin x = sin θ x = nπ + (-1)nθ, where θ ∈ [-π/2, π/2]

cos x = cos θ x = 2nπ ± θ, where θ ∈ (0, π]

tan x = tan θ x = nπ + θ, where θ ∈ (-π/2 , π/2]

sin2 x = sin2 θ x = nπ ± θ

cos2 x = cos2 θ x = nπ ± θ

tan2 x = tan2 θ x = nπ ± θ

MARK ME AS BRAINLISIT

✌✌✌