Найдите корень уравнения.корень 5-2x=7

5-2х=7-2х=7-5-2х=2

х=-2: 2

х=-1 

5-2x=7

-2x=7-5

-2x=2

x=-1

а)

[tex]2 \sin( \alpha ) + 2(1 - \sin( \alpha )
)[/tex]

2sin(a)+2-2sin(a)=2

б)

[tex]1 - \sin( { \alpha }^{2} ) - \cos( { \alpha }^{2} )
[/tex]

1-1=0

[tex]3)\left \{ {{b_{5}=0,64 } \atop
{b_{2}=0,08 }} \right.\\\\: \left \{ {{b_{1}*q^{4}=0,64} \atop {b_{1}*q=0,08}} \right.\\ {3}=8\\\\q=\sqrt[3]{8}=2\\\\b_{1}=\frac{0,08}{2} =0,04\\\\s_{9} =\frac{b_{1}(q^{9}-1)   }{q-1}=\frac{0,04(2^{9} -1)}{2-1}=0,04(512-1)=0,04*511=20,44[/tex]