Question

ayuda con estas ecuaciones trigonométricas ​

Answer 1

Explicación paso a paso:

1) $\sin^{2}{x} - \cos^{2}{x} = \cos{x}$

$(1- \cos^{2}{x}) - \cos^{2}{x} = \cos{x} \\ 1- \cos^{2}{x} - \cos^{2}{x} = \cos{x} \\ 1- 2\cos^{2}{x} = \cos{x} \\ 2\cos^{2}{x} + \cos{x} - 1 = 0$

consideramos $t = \cos{x}$ y aplicamos fórmula cuadrática

$2t^{2} + t - 1 = 0$

$t = \frac{-b \: \pm \: \sqrt{b^{2}-4ac} }{2a}$

$t = \frac{-1 \: \pm \: \sqrt{1^{2}-4(2)(-1)} }{2(2)}$

$t = \frac{-1 \: \pm \: \sqrt{9} }{4}$

$t = \frac{-1 \: \pm \: 3 }{4}$

$t_{1} = \frac{-1 + 3}{4} = \frac{2}{4} = 0.5 \\ \\ t_{2} = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$

$\cos{x} = \[ \left \{ \begin{array}{r} 0.5 \\ -1 \end{array} \right . \]$

Evaluando cada respuesta:

$\cos{x} = 0.5 \\ x = \cos^{-1}{(0.5)} \\ x = 60$

$\cos{x} = -1 \\ x = \cos^{-1}{(-1)} \\ x = 180$

Por tanto: $x = \[ \left \{ \begin{array}{r} 60 \\ 180 \end{array} \right . \]$

2) $2\sin{\alpha} + \sqrt{3} = 0$

$2\sin{\alpha} = -\sqrt{3} \\ \sin{\alpha} = \frac{-\sqrt{3}}{2} \\ \sin{\alpha} = - \frac{\sqrt{3}}{2}$

$\alpha = \sin^{-1}{(- \frac{\sqrt{3}}{2} )}$

$\alpha = - 60$

Por tanto: $\alpha = - 60$

3) $\sin{2x} + \cos{x} = 0$

$2 \sin{x} \cos{x} + \cos{x} = 0 \\ \cos{x} ( 2\sin{x} + 1 ) = 0$

Evaluando:

$\cos{x} = 0 \\ x = \cos^{-1}{0} \\ \\ x = \[ \left \{ \begin{array}{r} 90+360 \cdot n \\ 270+360 \cdot n \end{array} \right . \]$

$2\sin{x} + 1 = 0 \\ 2\sin{x} = -1 \\ \sin{x} = -0.5 \\ x = \sin^{-1}{-0.5} \\ \\ x = \[ \left \{ \begin{array}{r} 3.665191+360 \cdot n \\ 5.759586+360 \cdot n \end{array} \right . \]$

Por tanto:

$x = \[ \left \{ \begin{array}{r} 90+360 \cdot n \\ 270+360 \cdot n \\ 3.665191+360 \cdot n \\ 5.759586+360 \cdot n \end{array} \right . \]$

Cuando n = 1:

$x = \[ \left \{ \begin{array}{r} 90 \\ 270 \\ 3.665191 \\ 5.759586 \end{array} \right . \]$