Question

Sinx(2sin^2(x)-1)+cos^2(2x)=0; [-1,6; 0,8]

Answer 1

-sinx(1-2sin²x)+(1-2sin²x)²=0
(1-2sin²x)(-sinx+1-2sin²x)=0
cos2x*(2sin²x+sinx-1)=0
cos2x=0⇒2x=π/2+πn,n∈z⇒x=π/4+πn/2,n∈z
n=-1⇒x=-π/2+π/4=-π/4∈[-1,6;0,8]
n=0⇒x=π/4∈[-1,6;08]
2sin²x+sinx-1=0
sinx=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒sinx=-π/2+2πk,k∈z
k=0⇒x=-π/2∈[-1,6;0,8]
a2=(-1+3)/4=1/2⇒sinx=1/2
x=π/6+2πm,m∈z
m=0⇒x=π/6∈[-1,6;0,8]
x=5π/6+2πc,c∈z
нет корней на данном промежутке
x={-π/2;-π/4;π/4}