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2sin(2θ)=cos(2θ),0<= θ<= 2pi

Solución

2 sin (2 θ) = cos (2 θ), 0 \leq θ \leq 2 π

Solución

θ = \frac{0.46364 \dots}{2}, θ = \frac{0.46364 \dots + π}{2}, θ = \frac{0.46364 \dots + 2 π}{2}, θ = \frac{0.46364 \dots + 3 π}{2}

+1

Grados

θ = 13.28252 \dots^{\circ}, θ = 103.28252 \dots^{\circ}, θ = 193.28252 \dots^{\circ}, θ = 283.28252 \dots^{\circ}

Pasos de solución

2 sin (2 θ) = cos (2 θ), 0 \leq θ \leq 2 π

Re-escribir usando identidades trigonométricas

2 sin (2 θ) = cos (2 θ)

Dividir ambos lados entre

cos (2 θ), cos (2 θ) \neq = 0

\frac{2 sin ( 2 θ )}{cos ( 2 θ )} = \frac{cos ( 2 θ )}{cos ( 2 θ )}

Simplificar

\frac{2 sin ( 2 θ )}{cos ( 2 θ )} = 1

Utilizar la identidad trigonométrica básica:

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

2 tan (2 θ) = 1

2 tan (2 θ) = 1

Dividir ambos lados entre

2

2 tan (2 θ) = 1

Dividir ambos lados entre

2

\frac{2 tan ( 2 θ )}{2} = \frac{1}{2}

Simplificar

tan (2 θ) = \frac{1}{2}

tan (2 θ) = \frac{1}{2}

Aplicar propiedades trigonométricas inversas

tan (2 θ) = \frac{1}{2}

Soluciones generales para

tan (2 θ) = \frac{1}{2}

tan (x) = a \Rightarrow x = arctan (a) + πn

2 θ = arctan (\frac{1}{2}) + πn

2 θ = arctan (\frac{1}{2}) + πn

Resolver

2 θ = arctan (\frac{1}{2}) + πn : θ = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

2 θ = arctan (\frac{1}{2}) + πn

Dividir ambos lados entre

2

2 θ = arctan (\frac{1}{2}) + πn

Dividir ambos lados entre

2

\frac{2 θ}{2} = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

Simplificar

θ = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

θ = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

θ = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

Soluciones para el rango

0 \leq θ \leq 2 π

θ = \frac{arctan ( \frac{1}{2} )}{2}, θ = \frac{arctan ( \frac{1}{2} ) + π}{2}, θ = \frac{arctan ( \frac{1}{2} ) + 2 π}{2}, θ = \frac{arctan ( \frac{1}{2} ) + 3 π}{2}

Mostrar soluciones en forma decimal

θ = \frac{0.46364 \dots}{2}, θ = \frac{0.46364 \dots + π}{2}, θ = \frac{0.46364 \dots + 2 π}{2}, θ = \frac{0.46364 \dots + 3 π}{2}

Ejemplos populares

tan (4 x) - 1 = 0

(cos(x)cot(x))/(1-sin(x))=3

\frac{cos ( x ) cot ( x )}{1 - sin ( x )} = 3

sin(2x)sin(x)-cos(x)=0

sin (2 x) sin (x) - cos (x) = 0

tan^2(x)=-1+2tan(-x)

tan^{2} (x) = - 1 + 2 tan (- x)

tan(2θ)-tan(θ)=0

tan (2 θ) - tan (θ) = 0