Question

Решите тригонометрическое уравнение : sin^2 x-3cos x-3=0.

Answer 1

1- cos^2x-3cosx-3=0
cos^2x+3cosx+2=0
cosx=y
y^2+3y+2=0
D=9-8=1
y1=(-3-1)÷2=-2
y2=(-3+1)÷2=-1
X1=arccos (-2)+2pn
X2=arccos (-1)+2pn

Answer 2

1-cos^2x-3cosx-3=0

-cos^2x-3cosx-2=0

cos^2x+3cosx+2=0

cosx=t

t^2+3t+2=0

D=9-8=1

t1=(-3+1)/2=-2/2=-1

t2=(-3-1)/2=-4/2=-2 не подходит т.к. [-1;1]

cosx=-1

x=п+2пn , n∈Z