Question

1. 4 cos^2x+12sinx+3=0

2. (sin x/2 - 1)(ctgx+√3)=0

Answer 1

1)

$<var>4-4sin^2x+12sinx+3=0</var>$ 

$<var>4sin^2x-12sinx-7=0</var>$ 

sin x=3,5 или $<var>sinx=-1/2</var>$

реш.нет                 $<var>x= (-1)^{k+1}\frac{\pi}{6}+\pi*k</var>$

Ответ   $<var>x= (-1)^{k+1}\frac{\pi}{6}+\pi*k</var>$

2)

$<var>sin\frac{x}{2}=1</var>$ или $<var>ctg\ x=-\sqrt3</var>$ 

$<var>\frac{x}{2}=\frac{\pi}{2}+2\pi n</var>$ или $<var>x=\frac{5\pi}{6}+\pi m</var>$

x=П+4Пn

Ответ  П+4Пn;  $<var>\frac{5\pi}{6}+\pi m</var>$