Подробное решение с отбором корней

a) x=π/4+kπ/2, k∈Z

b) x={-13π/4; -11π/4; -9π/4;}

Объяснение:

a) √(0,5+sin²x)+cos2x=1

√(0,5+sin²x)+1-2sin²x=1

√(0,5+sin²x)=2sin²x

(√(0,5+sin²x))²=(2sin²x)²

0,5+sin²x=4sin⁴x

4sin⁴x-sin²x-0,5=0

sin²x=t, 0≤t≤1

4t²-t-0,5=0

D=1+8=9=3²

1) t=(1-3)/8=-1/4<0

2) t=(1+3)/8=1/2

sin²x=0,5

2sin²x=1

1-2sin²x=0

cos2x=0

2x=π/2+kπ

x=π/4+kπ/2, k∈Z

b) x∈[-3,5π; -2π]⇒x=

-3,5π≤π/4+kπ/2≤-2π

-3,5π-π/4≤kπ/2≤-2π-π/4

-7π-π/2≤kπ≤-4π-π/2

-7,5π≤kπ≤-4,5π

-7,5≤k≤-4,5, k∈Z⇒k={-7; -6; -5}

x=π/4+kπ/2,

x={-13π/4; -11π/4; -9π/4;}