Найти корни уравнения на [tex]( \pi ; \frac{3 \pi }{2} ][/tex]
[tex]cos(3x- \frac{ \pi }{2} )= \frac{1}{2} [/tex]
[tex]cos(3x- \frac{ \pi }{2} )= \frac{1}{2} [/tex]
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Verified answer
Cos(3x - π/2) = 1/2sin3x = 1/2
3x = (-1)^n*arcsin(1/2) + πk,k∈Z
3x = (-1)^(n)(π/6) + πk, k∈z
x = (-1)^(n)(π/18) +πk/3, n∈Z
cos(π/2 - 3х) = sin3x