Rozwiąż równanie:
2sin²x = 2 + cosx
2(1-cos²x) -cosx -2 =0
2 -2cos²x -cosx -2 =0
-2cos²x -cosx =0
-cosx (cosx -1) =0
cosx = 0 ∨ cosx-1 =0
x= ½π +kπ cosx = 1
x= 2kπ , k∈C
sin²x+cos²x=1
sin²x=1-cos²x
2(1-cos²x)=2+cosx
2-2cos²x=2+cosx
-2cos²x-cosx=0
cosx----t wprowadzamy zmienną t
-2t²-t=0
Δ=(-1)²-4*(-2)*0
Δ=1-0=1
√Δ=1
t₁=1-1/-4=0/-4=0
t₂=1+1/-4=2/-4=-1/2
cosx=0 lub cosx=-1/2
-π/2-kπ i π/2+kπ -π/3-2kπ i π/3+2kπ
-5/3-2kπ i 5/3+2kπ
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2sin²x = 2 + cosx
2(1-cos²x) -cosx -2 =0
2 -2cos²x -cosx -2 =0
-2cos²x -cosx =0
-cosx (cosx -1) =0
cosx = 0 ∨ cosx-1 =0
x= ½π +kπ cosx = 1
x= 2kπ , k∈C
sin²x+cos²x=1
sin²x=1-cos²x
2(1-cos²x)=2+cosx
2-2cos²x=2+cosx
-2cos²x-cosx=0
cosx----t wprowadzamy zmienną t
-2t²-t=0
Δ=(-1)²-4*(-2)*0
Δ=1-0=1
√Δ=1
t₁=1-1/-4=0/-4=0
t₂=1+1/-4=2/-4=-1/2
cosx=0 lub cosx=-1/2
-π/2-kπ i π/2+kπ -π/3-2kπ i π/3+2kπ
-5/3-2kπ i 5/3+2kπ