3 + 2cos²x - 1 + 3√2cosx = 0
2cos²x + 3√2cosx + 2 = 0
D = 18 - 16 = 2
cosx₁ = (-3√2 - √2)/4 = -√2 < -1 - не подходит
cosx₂ = (-3√2 + √2)/4 = -√2/2
x₂ = +- 3π/4 + 2πk, k∈Z
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3 + 2cos²x - 1 + 3√2cosx = 0
2cos²x + 3√2cosx + 2 = 0
D = 18 - 16 = 2
cosx₁ = (-3√2 - √2)/4 = -√2 < -1 - не подходит
cosx₂ = (-3√2 + √2)/4 = -√2/2
x₂ = +- 3π/4 + 2πk, k∈Z
Verified answer
3+cos2x+3√2cosx=03+cos²x-sin²x+3√2cosx=0
3+cos²x-1+cos²x+3√2cosx=0
2cos²x+3√2cosx+2=0
cosx=t€[-1;1]
2t²+3√2t+2=0
D=(3√2)²-16=18-16=2
t=(-3√2±√2)/4
t1=-√2;t2=-√2/2
1)cosx=-√2;x€∅
2)cosx=-√2/2
x=±arccos(-√2/2)+2πk
x=±(π-π/4)+2πk
x=±3π/4+2πk;k€Z