Question

Решите уравнения(по тригонометрии):
1. 4$sqrt{3}$cosx-4sinx=2$sqrt{3}$$cos x^{2}$-sin2x
2. $4- cos^{2}3x=3sin^{2}3x+2sin6x$
3.$4 sqrt{3}sinx-sin2x=2 sqrt{3}sin^{2} x-4cosx$
4.$frac{ left[sinx]}{sinx} -2=2cosx$

Answer 1

1
8(√3/2*cosx-1/2*sinx)-4cosx(√3/2cosx-1/2*sinx)=0
cos(x+π/6)*(8-4cosx)=0
cos(x+π/6)=0⇒x+π/6=π/2+πn,n∈z⇒x=π/3+πn,n∈z
8-4cosx=0⇒cosx=2>1 нет решения
2
4sin²3x+4cos²3x-cos²3x-3sin²3x-6sin3xcos3x=0/cos²3x
tg²3x-6tgx+3=0
tg3x=a
a²-6a+3=0
D=36-12=24
a1=(6-2√6)/2=3-√6⇒tg3x=3-√6⇒3x=arctg(3-√6)+πn,n⇒z⇒
x=1/3*arctg(3-√6)+πn/3,n∈z
a2=(6+2√6)/2=3+√6⇒tg3x=3+√6⇒3x=arctg(3+√6)+πk,k⇒z⇒
x=1/3*arctg(3+√6)+πk/3,k∈z
3
8(√3/2sinx+1/2cosx)-4sinx(√3/2sinx+1/2cosx)=0
cos(x-π/6)*(8-4sinx)=0
cos(x-π/6)=0⇒x-π/6=π/2+πn,n∈z⇒x=2π/3+πn,n∈z
8-4sinx=0⇒sinx=2>1 нет решения
4
1)sinx>0⇒x∈(2πn;π+2πn,n∈z)
sinx/sinx-2=2cosx
2cosx=-1
cosx=-1/2
x=2π/3+2πn,n∈z
2)sinx≤0⇒x∈[π+2πn;2π+2πn,n∈z]
-sinx/sinx-2=2cosx
2cosx=-3
cosx=-1,5<-1 нет решения