Question

$(4sin^2x + 12sinx - 7) * sqrt{-9cosx} =0$

ПОМОГИТЕЕЕ

 

 

Answer 1

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

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

x 1  не принадлежит

-9cosx=0

$<var>x = frac{pi}{2}+{pi}n$

x 1  не принадлежит

-9cosx=0

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

x 1  не принадлежит

-9cosx=0

$<var>x = frac{pi}{2} {pi}n$

Ответ $</var>frac{pi}{2}+{pi}n$\$<var></var> x_{2} = frac{5pi}{6}+2{pi}n\ frac{pi}{2}+2{pi}n" title="<var>x = frac{pi}{2}+{pi}n" /></var></var></var></p> <p>Ответ <var>[tex]</var>frac{pi}{2}+{pi}n$\$<var></var> x_{2} = frac{5pi}{6}+2{pi}n\ frac{pi}{2}+2{pi}n" alt="<var>x = frac{pi}{2}+{pi}n" /></var></var></var></p> <p>Ответ <var>[tex]</var>frac{pi}{2}+{pi}n$\[tex] x_{2} = frac{5pi}{6}+2{pi}n\ frac{pi}{2}+2{pi}n" />