Question

Ayuda le doy corona al que responda bien

Answer 1

Respuesta:

Explicación paso a paso:

1.

[tex]Sen (30) = Cos (3x)[/tex]

[tex]0.5 = Cos (3x)[/tex]

[tex](3x) = ArcCos ( 0.5 )[/tex]

[tex]3x = 60^{0}[/tex]

[tex]x = \frac{60^{0} }{3}[/tex]

Respuesta:

[tex]x = 20^{0}[/tex]

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2.

[tex]Tg ( 4x -20) = Ctg(5x+20)[/tex]

[tex]\frac{Sen(4x-20)}{Cos(4x-20)} =\frac{Cos(5x+20)}{Sen(5x+20)}[/tex]

[tex]Sen ( 4x-20) * Sen(5x+20) = Cos (4x-20) * Cos (5x+20)[/tex]

Aplicando la fórmula de transformación.

[tex]\frac{1}{2} [ Cos ( 4x-20-5x-20)-Cos ( 4x-20+5x+20) = \frac{1}{2} [ Cos (4x-20-5x-20)+Cos(4x-20+5x+20)][/tex]

[tex]\frac{1}{2} [ Cos( -x-40) -Cos ( 9x) ] = \frac{1}{2} [ Cos ( -x -40) + Cos ( 9 x ) ][/tex]

[tex]2Cos( -x-40) -2Cos(9x) = 2 Cos ( -x-40) + Cos (9x)[/tex]

[tex]2Cos (-x-40) -2Cos ( -x -40 ) = 2Cos (9x) + 2Cos (9x)[/tex]

[tex]0 = 4Cos (9x )[/tex]

[tex]Arc Cos ( 0) = 9x[/tex]

[tex]90^{0} = 9x[/tex]

[tex]x = \frac{90^{0} }{9}[/tex]

RESPUESTA:

     [tex]x= 10^{0}[/tex]

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3.

[tex]\sqrt{Tg(53)+4} = \sqrt{1.327+4} = \sqrt{5.327} = 2.31[/tex]

RESPUESTA:

    [tex]5.31[/tex]