a) sin4x=-4/5 b)sin(x/2-pi/3)=1 c)sin(pi/6+2x)=-1 d)sin(25⁰+ 2x)=-1/2 Có thể giúp mình với được không ạ ,mình thật sự rất cảm ơn????

Đáp án:

Giải thích các bước giải:

 a) `sin\ 4x=- 4/5`

`⇔ sin\ 4x=arcsin\ (- 4/5)`

`⇔` \(\left[ \begin{array}{l}4x=arcsin\ (-\dfrac{4}{5})+k2\pi\ (k \in \mathbb{Z})\\4x=\pi-arcsin\ (-\dfrac{4}{5})+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=\dfrac{arcsin\ (-\dfrac{4}{5})}{4}+k\dfrac{\pi}{2}\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{4}-\dfrac{arcsin\ (-\dfrac{4}{5})}{4}+k\dfrac{\pi}{2}\ (k \in \mathbb{Z})\end{array} \right.\) 

Vậy `S={x=\frac{arcsin\ (-\frac{4}{5})}{4}+k\frac{\pi}{2}\ (k \in \mathbb{Z});x=\frac{\pi}{4}-\frac{arcsin\ (-\frac{4}{5})}{4}+k\frac{\pi}{2}\ (k \in \mathbb{Z})}`

b) `sin (\frac{x}{2}-\frac{\pi}{3})=1`

`⇔ sin (\frac{x}{2}-\frac{\pi}{3})=sin \frac{\pi}{2}`

`⇔ \frac{x}{2}=\frac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})`

`⇔ x=\frac{5\pi}{3}+k4\pi\ (k \in \mathbb{Z})`

Vậy `S={\frac{5\pi}{3}+k4\pi\ (k \in \mathbb{Z})}`

c) `sin (\frac{\pi}{6}+2x)=-1`

`⇔ sin (\frac{\pi}{6}+2x)=sin (-\frac{\pi}{2})`

`⇔ \frac{\pi}{6}+2x=-\frac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})`

`⇔ 2x=-\frac{2\pi}{3}+k2\pi\ (k \in \mathbb{Z})`

`⇔ x=-\frac{\pi}{3}+k\pi\ (k \in \mathbb{Z})`

Vậy `S={-\frac{\pi}{3}+k\pi\ (k \in \mathbb{Z})}`

d) `sin (25^{0}+2x)=-\frac{1}{2}`

`⇔ sin (25^{0}+2x)=sin (-30^{0})`

`⇔` \(\left[ \begin{array}{l}25^{0}+2x=-30^{0}+k360^{0}\ (k \in \mathbb{Z})\\25^{0}+2x=180^{0}-30^{0}+k360^{0}\ (k \in \mathbb{Z})\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}2x=-55^{0}+k360^{0}\ (k \in \mathbb{Z})\\2x=125^{0}+k360^{0}\ (k \in \mathbb{Z})\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=\dfrac{-55^{0}}{2}+k180^{0}\ (k \in \mathbb{Z})\\x=\dfrac{125^{0}}{2}+k180^{0}\ (k \in \mathbb{Z})\end{array} \right.\) 

Vậy `S={\frac{-55^{0}}{2}+k180^{0}\ (k \in \mathbb{Z});\frac{125^{0}}{2}+k180^{0}\ (k \in \mathbb{Z})}`