решите уравнения (3ctg3x+√3)(tg4x+1)=0 ctg(x/2+пи/4)-√3=0 tgx=1 x ∈[0;2пи]. Created by darya280797. algebra-ru

1. (3ctg3x+√3)(tg4x+1)=0(3ctg3x+√3) = 0ctg3x = -/33x = π/3 + πn, n ∈ Zx = π/6 + π/3n, n ∈ Z(tg4x+1)=0tg4x = -1 4x = π/4 + πn, n ∈ Z;x = π/16 + π/4n, n ∈ Z2. ctg(x/2+пи/4)-√3=0ctg(x/2+пи/4) = x/2 + π/4 = π/6 + πn, n ∈ Z;x = π/3 - π/2 + 2πn,n ∈ Z'x = -π/6 + 2πn, n ∈ Z;3.tgx=1 x ∈[0;2пи]x = π/4 + πn,n ∈ Z;Поскольку x ∈[0;2пи] тоx = π/4 + πn, n = 0,1;

4)-√3=0 tgx=1 x ∈[0;2пи]...

darya280797 @darya280797

July 2022 1 3 Report

решите уравнения
(3ctg3x+√3)(tg4x+1)=0
ctg(x/2+пи/4)-√3=0
tgx=1 x ∈[0;2пи]

Answers & Comments

NoNamegirl 1. (3ctg3x+√3)(tg4x+1)=0
(3ctg3x+√3) = 0
ctg3x = - $\sqrt{3}$ /3
3x = π/3 + πn, n ∈ Z
x = π/6 + π/3n, n ∈ Z
(tg4x+1)=0
tg4x = -1
4x = π/4 + πn, n ∈ Z;
x = π/16 + π/4n, n ∈ Z

2. ctg(x/2+пи/4)-√3=0
ctg(x/2+пи/4) = $\sqrt{3}$
x/2 + π/4 = π/6 + πn, n ∈ Z;
x = π/3 - π/2 + 2πn,n ∈ Z'
x = -π/6 + 2πn, n ∈ Z;

3.tgx=1 x ∈[0;2пи]
x = π/4 + πn,n ∈ Z;
Поскольку x ∈[0;2пи] то
x = π/4 + πn, n = 0,1;

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