Question

ПОМОГИТЕЕЕ

Answer 1

$\left \{ {{4sin^2x + 12sinx - 7 =0} \atop {-9cosx>0}} \right\\ \left \{ {{4(sinx -0.5)(sinx +3.5) =0} \atop {cosx<0}} \right$

$x_{1} = \frac{\pi}{6}+2{\pi}n\\ x_{2} = \frac{5\pi}{6}+2{\pi}n\\ \frac{\pi}{2}+2{\pi}n<x<\frac{3{\pi}}{2}+2{\pi}n< var="">$

x 1  не принадлежит

-9cosx=0

$x = \frac{\pi}{2}+{\pi}n$

Ответ <var>$\frac{\pi}{2}+{\pi}n$\\<var>$x_{2} = \frac{5\pi}{6}+2{\pi}n\\ \frac{\pi}{2}+2{\pi}n$</var></var>