Question

7 задание ....................

Answer 1

cos²x-sin²x-cos²x-√2sinx=0

-sinx*(sinx+√2)=0;  sinx=-√2,∅, т.к.  -1 ≤sinx≤1

sinx=0

x=πn; n∈Z

-π≤πn≤π

-1≤n≤1

n=-1; x=-π

n=0; х=0

n=1; х=π

Answer 2

Ответ:

$cos2x-cos^2x-\sqrt2sinx=0\\\\(cos^2x-sin^2x)-cos^2x-\sqrt2sinx=0\\\\-sin^2x-\sqrt2sinx=0\\\\-sinx(sinx+\sqrt2)=0\\\\a)\ \ sinx=0\ \ ,\ \ \underline {x=\pi n\ ,\ n\in Z}\\\\b)\ \ sinx+\sqrt2=0\ \ \to \ \ sinx=-\sqrt2<-1\ \ \to \ \ \ x\in \varnothing \ ,\\\\t.k.\ \ |sinx|\leq 1\ ,\\\\c)\ \ x\in [-\pi \, ;\, \pi \ ]:\ \ \underline {x_1=-\pi \ ,\ \ x_2=0\ ,\ \ x_3=\pi }\ .\\\\Otvet:\ \ x=\pi n\ ,\ n\in Z\ \ ;\ \ \ x_{1,2,3}=-\pi ,\ 0\ ,\ \pi \in [-\pi \, ;\, \pi \ ]\ .$