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Qui peux m'aider à résoudre
1/ sin ( 2x - π/4 ) = sin ( π/3 - x )
2/ sin 2x = cos ( x - 2π/3 )
Merci
Qui peux m'aider à résoudre
1/ sin ( 2x - π/4 ) = sin ( π/3 - x )
2/ sin 2x = cos ( x - 2π/3 )
Merci
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1/ sin ( 2x - π/4 ) = sin ( π/3 - x ){2x-π/4=π/3-x+2kπ
{2x-π/4=π-π/3+x+2kπ
donc
{3x=7π/12+2kπ
{x=11π/12+2kπ
donc
{x=7π/36+2kπ/3
{x=11π/12+2kπ
2/ sin 2x = cos ( x - 2π/3 )
cos(π/2-x)=cos(x-2π/3)
{π/2-x=x-2π/3+2kπ
{π/2-x=-x+2π/3+2kπ
donc
{-2x=-7π/6+2kπ
{impossible
donc
{x=7π/12+kπ
sin(U) = sin(V) ⇔ U = V + 2kπ ou U = π - V + 2kπ
U = 2 x - π/4
V = π/3 - x
⇔ 2 x - π/4 = π/3 - x +2kπ
2 x + x = π/3 + π/4 + 2kπ
3 x = 7π/12 + 2kπ
x = 7π/36 + 2kπ/3
U = π - V + 2kπ ⇔ 2 x - π/4 = π - (π/3 - x) + 2kπ
2 x - π/4 = π - π/3 + x) + 2kπ
x = 2π/3 + π/4 + 2kπ = 11π/12 + 2kπ
pour l'ex 2 vous faite la même démarche