Question

Решите уравнения:
1) tg(x-3pi/4)+1=0
2) ctg(3,5pi+x)=0
3) 3sinx-8=sqrt2

Answer 1

$1) \ tg \bigg(x - \dfrac{3\pi}{4} \bigg) + 1 = 0$

$tg \bigg(x - \dfrac{3\pi}{4} \bigg) = -1$

$x - \dfrac{3\pi}{4} = arctg(-1)$

$x - \dfrac{3\pi}{4} = -\dfrac{\pi}{4}$

$x - \dfrac{3\pi}{4} = -\dfrac{\pi}{4} + \pi n, n \in Z$

$x = -\dfrac{\pi}{4} + \dfrac{3\pi}{4} + \pi n, n \in Z$

$x = \dfrac{\pi}{2} + \pi n, n \in Z$

$2) \ ctg (3,5 \pi + x) = 0$

$-tg \ x = 0$

$tg \ x = 0$

$x = \pi n, n \in Z$

$3) \ 3sin \ x - 8 = \sqrt{2}$

ОДЗ: $x \in [-1; 1]$

$3sin \ x = \sqrt{2} + 8$

$sin \ x \approx \dfrac{1,41 + 8}{3} = \dfrac{9,41}{3} \approx 3,14$

$\oslash$ (Нет решений).

Answer 2

1)tg(x-3π/4)+1=0
tg(x-3π/4)=-1
x-3π/4=-π/4+πk
x=3π/4-π/4+πk
x=2π/4+πk
x=π/2+πk
2)ctg(3,5π+x)=0
3,5π+x=π/2+πk
x=-3,5π+0,5π+πk
x=-3π+πk

3)3sinx-8=√2
sinx€[-1;1]
3sinx=8+√2
sinx=((8+√2)/3)>1
x€∅