sin(6x - π/3)=sin(2x + π/4)
ответ: 7π/4 +(π/2)*n n ∈ ℤ . π/6 +(π/4)* k , k ∈ ℤ.
Объяснение:
sin(6x - π/3) = sin(2x + π/4)
(6x - π/3) - (2x+ π/4) = 2πn ⇒ x = 7π/4 +(π/2)*n , n ∈ ℤ.
(6x - π/3) + (2x+ π/4) = π+ 2πk ⇒ x = π/6 +(π/4)* k , k ∈ ℤ.
P.S. * * * * * * * * * * * * * * * * * * * * * * * * * * *
sinα=sinβ ⇔sinα- sinβ =0 ⇔2(sin(α- β)/2) )* ( cos(α+ β)/2) )=0.
sin(α - β)/2 =0 ⇒ (α- β)/2 = π*n ⇔ α- β = 2π*n ;
cos(α+ β)/2 =0 ⇒(α + β)/2 = π/2 +π*k⇔ α + β =π +2π*k . || (2k+1)π ||
а) (x + 2) (x - 2) - (x + 4) (x - 4) + (x - 5) (x + 5)= (х²—4)–(х²–16)+(х²–25)= х²–4–х²+16+х²–25= х²–13
б) (y - 3) (y + 3) + (2y - 1) (2y + 1) - (y + 1) (y - 1)= (у²–9)+(4у²–1)–(у²–1)= у²–9+4у²–1–у²+1= 4у²–9
в) (t - 1) (t + 1) (t² + 1)= (t²–1)(t²+1)= t⁴–1
r) (u² + v²) (u - v) (u + v)= (u³–u²v+uv²–v³)(u+v)= u⁴–u³v+u²v²–uv³+u³v–u²v²+uv³–v⁴=u⁴–v⁴
г) (a + x - z) (a + x + 2)= а²+ах+2а+ах+х²+2х–аz–xz–2z= a²+x²+2a+2x–2z+2ax–аz–xz
д) (x³ + x + 2) (x² + x - 2)= (х^5)+х⁴–2х³+х³+х²–2х+2х²+2х–4=(х^5)+х⁴–х³+3х²–4
е) (a + b + c²) (b + c²- a)= аb+ac²–a²+b²+bc²–ab+bc²+c⁴–ac²= –a²+b²+2bc²+c⁴
ж) (c² + ab - d²) (ab - c² + d²)= abc²–c⁴+c²d²—a²b²–abc²+abd²–abd²+d²c²–d⁴=–c⁴+2c²d²–a²b²–d⁴
Треугольник CEF прямоугольный.. Тогда угол ECF = 90-65=25