2cos²x-sinx-1=0
2(1-sin²x)-sinx-1=0
-2sin²x-sinx+1=0
2sin²x+sinx-1=0
sinx=y
2y²+y-1=0
y₁=½,y₂=-1
1) sinx=½;x=(-1)ᵏ•(π/6)+πk,k∈Z
2) sinx=-1;x=-π/2+2πn,n∈Z.
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2cos²x-sinx-1=0
2(1-sin²x)-sinx-1=0
-2sin²x-sinx+1=0
2sin²x+sinx-1=0
sinx=y
2y²+y-1=0
y₁=½,y₂=-1
1) sinx=½;x=(-1)ᵏ•(π/6)+πk,k∈Z
2) sinx=-1;x=-π/2+2πn,n∈Z.