1)sin x/3= 6/5 2)tg (4x +pi /3) = - √3 3)sin (3x +pi/ 6) =-√ 3/2 4)cos²x-2cosx*sinx=0...

Question 1

1)sin x/3= 6/5
2)tg (4x +pi /3) = - √3
3)sin (3x +pi/ 6) =-√ 3/2
4)cos²x-2cosx*sinx=0
5)sin( x/2) ≥√2/2
6)cos3x≥√3/2

до завтра пожалйста

Question 2

Решите задачу:

$1)\; \; sin \frac{x}{3} = \frac{6}{5}\; \;net\; reshenij ,\; t.k.\; |sinx| \leq 1\\\\2)\; \; tg(4x+ \frac{\pi}{3})=-\sqrt3\\\\4x+ \frac{\pi}{3}=- \frac{\pi}{3} +\pi n,\; \; \; 4x =-\frac{2\pi}{3}+\pi n\; ,\; n\in Z\\\\x=-\frac{\pi}{6}+\frac{\pi n}{6}\; ,\; n\in Z\\\\3)\; \; sin(3x+ \frac{\pi }{6} )=-\frac{\sqrt3}{2}\\\\3x+ \frac{\pi}{6}=(-1)^{n}(- \frac{\pi}{3} )+\pi n=(-1)^{n+1}\frac{\pi}{3}+\pi n\; ,\; n\in Z\\\\3x=-\frac{\pi}{6}+(-1)^{n+1} \frac{\pi}{3}+\pi n$

$x=(-1)^{n+1} \frac{\pi}{9}- \frac{\pi}{18}+\pi n,\; n\in Z$

$4)\; \; cos^2x-2cosx\cdot sinx=0\\\\cosx(cosx-2sinx)=0\\\\cosx=0\; ,\; \; x= \frac{\pi }{2}+\pi n,\; n\in Z\\\\cosx-2sinx=0\; |:cosx\ne 0\\\\-2tgx+1=0\; ,\; \; tgx= \frac{1}{2} \\\\x=arctg \frac{1}{2}+\pi n,\; n\in Z\\\\5)\; \; sin\frac{x}{2} \geq \frac{\sqrt2}{2}\\\\ \frac{\pi}{4}+2\pi n \leq \frac{x}{2} \leq \frac{3\pi }{4}+2\pi n\; ,\; n\in Z\\\\ \frac{\pi }{2}+\pi n \leq x \leq \frac{3\pi }{2}+\pi n,\; n\in Z\\\\6)\; \; cos3x \geq \frac{\sqrt3}{2}$

$-\frac{\pi}{6} +2\pi n\leq 3x \leq \frac{\pi}{6}+2\pi n,\; n\in Z$

$- \frac{\pi }{6}+ \frac{2\pi}{3} \leq x \leq \frac{\pi}{18}+ \frac{2\pi n}{3}\; ,\; n\in Z$

Question 3

СПАСИБО БОЛЬШОЕ !!!