paulineoleron Tan^2 (x+pi/8) = 1/3
sin^2 (x+pi/8) ÷ cos^2 (x+pi/8) = 1/3
sin^2 (x+pi/8) = cos^2 (x+pi/8) × 1/3

or sin^2 (x) + cos^2 (x) = 1
sin^2 (x) = 1 - cos^2 (x)

1 - cos^2 (x+pi/8) = cos^2 (x+pi/8) x 1/3
1 = 4/3 x cos^2 (x+pi/8)
cos^2 (x+pi/8) = 3/4
cos (x+pi/8) = (racine de 3) ÷ 2

sur lR
x + pi/8 = pi/6 +2kpi x + pi/8 = - pi/6 + 2kpi
x = pi/24 + 2kpi x = -7pi/24 + 2kpi

aymanemaysae Bonsoir ;

Exercice n° 3 .

tan²(x + π/8) = 1/3 = (1/√3)² = tan²(π/6) ,

donc : tan²(x + π/8) - tan²(π/6) = 0 ,

donc : (tan(x + π/8) - tan(π/6))(tan(x + π/8) + tan(π/6)) = 0 ,

donc : tan(x + π/8) - tan(π/6) = 0  ou tan(x + π/8) + tan(π/6) = 0 .

Résolution de tan(x + π/8) - tan(π/6) = 0 .

tan(x + π/8) - tan(π/6) = 0 ,
donc : tan(x + π/8) = tan(π/6) ,
donc : x + π/8 = π/6 + kπ ; k ∈ Z ,
donc : x = π/6 - π/8 + kπ ; k ∈ Z ,
donc : x = π/24 + kπ ; k ∈ Z ,
donc l'ensemble des solutions de l'équation (tan(x + π/8) - tan(π/6) = 0) ,
est : S1 = {π/24 + kπ ; k ∈ Z} .

Résolution de tan(x + π/8) + tan(π/6) = 0 .

tan(x + π/8) + tan(π/6) = 0 ,
donc : tan(x + π/8) = - tan(π/6) = tan(- π/6) ,
donc : x + π/8 = - π/6 + kπ ; k ∈ Z ,
donc : x = - π/6 - π/8 + kπ ; k ∈ Z ,
donc : x = - 7π/24 + kπ ; k ∈ Z ,
donc l'ensemble des solutions de l'équation (tan(x + π/8) + tan(π/6) = 0) ,
est : S2 = {- 7π/24 + kπ ; k ∈ Z} .

Conclusion : La solution de l'équation (tan²(x + π/8) = 1/3)
est S = S1 ∪ S2 = {π/24 + kπ ; k ∈ Z} ∪ {- 7π/24 + kπ ; k ∈ Z} .