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Popular >

2sin(2θ)=cos(2θ+30)

Solución

2 sin (2 θ) = cos (2 θ + 3 0^{\circ})

Solución

θ = \frac{0.33347 \dots}{2} + \frac{18 0 ^{\circ} n}{2}

radianes

θ = \frac{0.33347 \dots}{2} + \frac{π}{2} n

Pasos de solución

2 sin (2 θ) = cos (2 θ + 3 0^{\circ})

Re-escribir usando identidades trigonométricas

2 sin (2 θ) = cos (2 θ + 3 0^{\circ})

Re-escribir usando identidades trigonométricas

cos (2 θ + 3 0^{\circ})

Utilizar la identidad de suma de ángulos:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (2 θ) cos (3 0^{\circ}) - sin (2 θ) sin (3 0^{\circ})

Simplificar

cos (2 θ) cos (3 0^{\circ}) - sin (2 θ) sin (3 0^{\circ}) : \frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

cos (2 θ) cos (3 0^{\circ}) - sin (2 θ) sin (3 0^{\circ})

Simplificar

cos (3 0^{\circ}) : \frac{3}{2}

cos (3 0^{\circ})

Utilizar la siguiente identidad trivial:

cos (3 0^{\circ}) = \frac{3}{2}

cos (x)

tabla de valores periódicos con

36 0^{\circ} n

intervalos:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2} cos (2 θ) - sin (3 0^{\circ}) sin (2 θ)

Simplificar

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

Utilizar la siguiente identidad trivial:

sin (3 0^{\circ}) = \frac{1}{2}

tabla de valores periódicos con

36 0^{\circ} n

intervalos:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= \frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

= \frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

2 sin (2 θ) = \frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

2 sin (2 θ) = \frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

Restar

\frac{3}{2} cos (2 θ) - \frac{1}{2} sin (2 θ)

de ambos lados

\frac{5}{2} sin (2 θ) - \frac{3}{2} cos (2 θ) = 0

Simplificar

\frac{5}{2} sin (2 θ) - \frac{3}{2} cos (2 θ) : \frac{5 sin ( 2 θ ) - 3 cos ( 2 θ )}{2}

\frac{5}{2} sin (2 θ) - \frac{3}{2} cos (2 θ)

Multiplicar

\frac{5}{2} sin (2 θ) : \frac{5 sin ( 2 θ )}{2}

\frac{5}{2} sin (2 θ)

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{5 sin ( 2 θ )}{2}

= \frac{5 sin ( 2 θ )}{2} - \frac{3}{2} cos (2 θ)

Multiplicar

\frac{3}{2} cos (2 θ) : \frac{3 cos ( 2 θ )}{2}

\frac{3}{2} cos (2 θ)

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 cos ( 2 θ )}{2}

= \frac{5 sin ( 2 θ )}{2} - \frac{3 cos ( 2 θ )}{2}

Aplicar la regla

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{5 sin ( 2 θ ) - 3 cos ( 2 θ )}{2}

\frac{5 sin ( 2 θ ) - 3 cos ( 2 θ )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

5 sin (2 θ) - 3 cos (2 θ) = 0

Re-escribir usando identidades trigonométricas

5 sin (2 θ) - 3 cos (2 θ) = 0

Dividir ambos lados entre

cos (2 θ), cos (2 θ) \neq = 0

\frac{5 sin ( 2 θ ) - 3 cos ( 2 θ )}{cos ( 2 θ )} = \frac{0}{cos ( 2 θ )}

Simplificar

\frac{5 sin ( 2 θ )}{cos ( 2 θ )} - 3 = 0

Utilizar la identidad trigonométrica básica:

\frac{sin ( x )}{cos ( x )} = tan (x)

5 tan (2 θ) - 3 = 0

5 tan (2 θ) - 3 = 0

Mover

3

al lado derecho

5 tan (2 θ) - 3 = 0

Sumar

3

a ambos lados

5 tan (2 θ) - 3 + 3 = 0 + 3

Simplificar

5 tan (2 θ) = 3

5 tan (2 θ) = 3

Dividir ambos lados entre

5

5 tan (2 θ) = 3

Dividir ambos lados entre

5

\frac{5 tan ( 2 θ )}{5} = \frac{3}{5}

Simplificar

tan (2 θ) = \frac{3}{5}

tan (2 θ) = \frac{3}{5}

Aplicar propiedades trigonométricas inversas

tan (2 θ) = \frac{3}{5}

Soluciones generales para

tan (2 θ) = \frac{3}{5}

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

2 θ = arctan (\frac{3}{5}) + 18 0^{\circ} n

2 θ = arctan (\frac{3}{5}) + 18 0^{\circ} n

Resolver

2 θ = arctan (\frac{3}{5}) + 18 0^{\circ} n : θ = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

2 θ = arctan (\frac{3}{5}) + 18 0^{\circ} n

Dividir ambos lados entre

2

2 θ = arctan (\frac{3}{5}) + 18 0^{\circ} n

Dividir ambos lados entre

2

\frac{2 θ}{2} = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

Simplificar

θ = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

θ = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

θ = \frac{arctan ( \frac{3}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

Mostrar soluciones en forma decimal

θ = \frac{0.33347 \dots}{2} + \frac{18 0 ^{\circ} n}{2}

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