Решите тригонометрическое уравнение: √3sinx+cosx=√2
√3sinx+cosx=√2
√3/2sinx+1/2cosx=√2/2sinπ/6sinx+cosπ/6cosx=√2/2cos(x-π/6)=√2/2 (x-π/6)=±arcos√2/2+2πn(n ∈ Z)x=±π/4+π/6 +2πn(n ∈ Z)x=5π/12+2πn(n ∈ Z)иx=-π/12+2πn(n ∈ Z)
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Answers & Comments
√3sinx+cosx=√2
√3/2sinx+1/2cosx=√2/2
sinπ/6sinx+cosπ/6cosx=√2/2
cos(x-π/6)=√2/2
(x-π/6)=±arcos√2/2+2πn(n ∈ Z)
x=±π/4+π/6 +2πn(n ∈ Z)
x=5π/12+2πn(n ∈ Z)
и
x=-π/12+2πn(n ∈ Z)