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Saires777
@Saires777
August 2022
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8) Прошу. Помогите!
a) sin 2x = [tex] \frac{1}{2} [/tex]
b) 1+2cosx=0
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HelperMath
Verified answer
А) sin2x = 1/2
2x = (-1)^k * π/6 + πk, k∈Z
x = (-1)^k * π/12 + πk/2, k∈Z
б) 1 + 2cosx = 0
2cosx = -1
cosx = -1/2
x = +/- 2π/3 + 2πk, k∈Z
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Answers & Comments
Verified answer
А) sin2x = 1/22x = (-1)^k * π/6 + πk, k∈Z
x = (-1)^k * π/12 + πk/2, k∈Z
б) 1 + 2cosx = 0
2cosx = -1
cosx = -1/2
x = +/- 2π/3 + 2πk, k∈Z