[tex]2sin^2x-3cosx-3=0[/tex]
помогите пожалуйста
cosx=t
2t^2+3t+1=0
D=1
t1=-1,
t2=-½,
cosx=-1, x=π+2πk, k∈Z,
cosx=-½, x=±arccos(-½)+2πk, x=±(π-arccos½)+2πk, x=±(π-π/3)+2πk, x=±2π/3+2πk, k∈Z
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Verified answer
cosx=t
2t^2+3t+1=0
D=1
t1=-1,
t2=-½,
cosx=-1, x=π+2πk, k∈Z,
cosx=-½, x=±arccos(-½)+2πk, x=±(π-arccos½)+2πk, x=±(π-π/3)+2πk, x=±2π/3+2πk, k∈Z