Question

2sin x -3cos x =4

2sin2x + 2cos2x +1 = 0

Answer 1

Відповідь:

1) x ∈ R

2) x ≈

{1.35878+kπ

{-0.573383+kπ

k ∈ Z

Пояснення:

1) 2sin(π+2kπ)-3cos(π+2kπ)=4

2sin(π)-3cos(π)=4

3=4

x ∈ R

2) 2sin(2x)+2cos(2x)+1=0

4sin(x)cos(x)+2-2sin(x)²-2sin(x)²=-1

4sin(x)cos(x)-4sin(x)²+3sin(x)²+3cos(x)²=0

4sin(x)cos(x)-sin(x)²+3cos(x)²=0

4tan(x)-tan(x)²+3=0

4t-t²+3=0

{tan(x)=2+√7

{tan(x)=2-√7

{x=arctan(2+√7)+kπ, k ∈ Z

{x=arctan(2-√7)+kπ, k ∈ Z