Matemáticas, pregunta formulada por guevaradiana197, hace 1 año

alguien me ayuda por favor​

Adjuntos:

Respuestas a la pregunta

Contestado por JainFelix
1

Explicación paso a paso:

a)

sen( \alpha ) =   \frac{ \sqrt{2} }{2}   \to \:  \alpha   = {45}^{0}

b)

2cos(2 \theta) -  \sqrt{3}  = 0

cos(2 \theta) =  \frac{ \sqrt{3} }{2}

arccos( \frac{ \sqrt{3} }{2} ) =  {30}^{0}

 2\theta  =  {30}^{0} \to \:  \theta =  {15}^{0}

c)

 \sqrt{3} tan( \frac{1}{3} t) = 1

tan( \frac{1}{3} t) =  \frac{1}{ \sqrt{3} }

arctan( \frac{1}{ \sqrt{3} } ) =  {60}^{0}

 \frac{1}{3} t =  {60}^{0}  \to \: t =  {180}^{0}

d)

2sen(3 \theta) +  \sqrt{2}  = 0

sen( 3\theta) =  -  \frac{ \sqrt{2} }{2}

arcsen(  - \frac{  \sqrt{2} }{2} ) =  -  {45}^{0}

3 \theta = - {45}^{0}  \to \:  \theta = - {15}^{0}

e)

sen( \theta +  \frac{\pi}{2} ) =  \frac{1}{2}

sen( \frac{2 \theta + \pi}{2} ) =  \frac{1}{2}

\pi =  {180}^{0}

arcsen( \frac{1}{2} ) =  {30}^{0}

2 \theta + {180}^{0}  =2(  {30}^{0} )

2 \theta =  -  {120}^{0}  \to \:  \theta =   - {60}^{0}

f)

cos( \alpha -  \frac{\pi}{3} ) =  - 1

 \frac{\pi}{3}  =  \frac{ {180}^{0} }{3}  =  {60}^{0}

arccos( - 1) =  {180}^{0}

 \alpha  - ( {60}^{0} ) =  {180}^{0}  \to \:  \alpha  =  {240}^{0}

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