Question

решите тригонометрические уравнения: 1) 2sin2x-3cosx=0 2)sin^2 4x-cos^2 4x=1 3)-5sinx-2cos^2x=1 4)cos(x-п/4)=-1

Answer 1

1) 4sinx*cosx-3cosx=0
cosx(4sinx-3)=0
cosx=0 или 4sinx-3=0
a) x=π/2+πk
б) sinx=3/4
x=(-1)^n*arcsin(3/4)+πn
n,k∈Z
2) -cos8x=-1
cos8x=1
8x=π/2+πk
x=π/16+πk/8,
k∈Z
4) cos(x-π/4)=-1
x-π/4=π+2πk
x=π+π/4+2πk
x=5π/4+2πk,
k∈Z

3) -5sinx-2cos^2x=1
$-5sinx-2(1-sin^2x)=1$
$-5sinx-2+2sin^2x=1$
$2sin^2x-5sinx-3=0$
Пусть sinx=a => a∈[-1; 1]
$2a^{2} -5a-3=0 \\ D=25+24=49=7^2 \\ a_1= \frac{5+7}{4}=3 \ \textgreater \ 1 \\ a_2= \frac{5-7}{4}=-1/2$
sinx=-1/2
$x=(-1)^k (-\pi /6)+ \pi k=(-1)^{k+1} \pi /6+ \pi k$,
k∈Z