Task/24988982
--------------------
Решить уравнение 2cos(x+π/3) +4sin(x+π/3) =5/2
-------
Решение: 2cos(x+π/3) +4sin(x+π/3) =5/2 ⇔2sin(x+π/3)+cos(x+π/3) =5/4.
---
* * * по методу дополнительного (вспомогательного) угла asinx +bcosx =√(a²+b²)sin(x +φ) , где φ =arctg(b/a) * * *
Сразу
sin( x+π/3+arctg(1/2) )=(√5)/4
x+π/3+arctg(1/2) = (-1)^n arcsin(√5)/4 +πn , n∈ℤ.
x = - π/3- arctg(1/2) + (-1)^n arcsin(√5)/4 +πn , n∈ℤ.
---ИЛИ после некоторого упрощения ---
2cos(x+π/3) +4sin(x+π/3) =5/2 ⇔
2*(cosx*cos(π/3) -sinx*sin(π/3) )+4*(sinx*cos(π/3) +cosx*sin(π/3) )=5/2 ⇔
cosx -√3sinx +2sinx+2√3cosx =5/2. ⇔
(2-√3)sinx +(1+2√3)cosx =5/2. ⇔
2√5sin(x+arctq(1+2√3)/(2-√3) =5/2 ⇔
sin(x+arctq(1+2√3)*(2+√3) ) = (√5)/4 и т.д .
* * * P.S * * *
2sin(x+π/3)+cos(x+π/3) =5/4.⇔
√(2²+1²) *( (2/√5)sin(x+π/3)+(1/√5)cos(x+π/3) ) =5/4.
обозначаем : cosφ =(2/√5) ⇒ sinφ =1/√5 ,φ =1/2.
√(2²+1²) sin(x+π/3+arctg(1/2) ) =5/4.