√сosx=√2sinx/2
ОДЗ
cosx≥0⇒x∈[-π/2+2πk;π/2+2πk,k∈z]
Возведем в квадрат
cosx=4sin²x/2
1-2sin²x/2=2sinx/2
4sin²x/2=1
sin²x/2=1/4
sinx/2=-1/2⇒x/2=-π/6+2πk U x=-5π/6+2πk,k∈z ⇒-π/3+4πk U x=-5π/3+4πk,k∈zx=+ОДЗ⇒х=-π/3+4πk,k∈z
sinx/2=1/2⇒x/2=π/6+2πk U x=5π/6+2πk,k∈z ⇒π/3+4πk U x=5π/3+4πk,k∈zx=+ОДЗ⇒х=π/3+4πk,k∈z
Ответ х=π/3+4πk,k∈z;x=-π/3+4πk,k∈z
(6cos²x-cosx-2)/√(-sinx)=0
ОДЗ sinx<0⇒x∈(π+2πk;2π+2πk,k∈z)<br>cosx=t
6t²-t-2=0
D=1+48=49
t1=(1-7)/12=-1/2⇒cosx=-1/2
x=-2π/3+2πk U x=2π/3+2πk +ОДЗ⇒х=4π/3+2π,k∈z
t2=(1+7)/12=2/3⇒cosx=2/3
x=-arccos2/3+2πk U x=arccos2/3+2πk,k∈z+ОДЗ⇒x=-arccos2/3+2πk,k∈z
Ответ х=4π/3+2π,k∈z;x=-arccos2/3+2πk,k∈z
5sinx+12cosx=12
10sin(x/2)cos(x/2)+12cos²(x/2)+12sin²(x/2)-12cos²(x/2)-12sin²(x/2)=0
10sin(x/2)cos(x/2)-24sin²(x/2)=0
2sin(x/2)*(5cos(x/2)-12sin(x/2))=0
sin(x/2)=0⇒x/2=πk⇒x=2πk,k∈z
5cos(x/2)-12sin(x/2)=0/cos(x/2)
5-12tg(x/2)=0
tg(x/2)=5/12
x/2=arctg5/12+πk
x=2arctg5/12+2πk,k∈z