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14.79m=((7.61_{0}^2sin(2θ_{0}))/g)

Solução

14.79 m = (\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} )}{g})

θ_{0} = \frac{arcsin ( 0.25538 \dots m g )}{2} + πn, θ_{0} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + πn

Passos da solução

14.79 m = (\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} )}{g})

Trocar lados

\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} )}{g} = 14.79 m

Multiplicar ambos os lados por

g

\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} )}{g} = 14.79 m

Multiplicar ambos os lados por

g

\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} ) g}{g} = 14.79 m g; g \neq = 0

Simplificar

57.9121 sin (2 θ_{0}) = 14.79 m g; g \neq = 0

57.9121 sin (2 θ_{0}) = 14.79 m g; g \neq = 0

Dividir ambos os lados por

57.9121; g \neq = 0

57.9121 sin (2 θ_{0}) = 14.79 m g; g \neq = 0

Dividir ambos os lados por

57.9121; g \neq = 0

\frac{57.9121 sin ( 2 θ _{0} )}{57.9121} = \frac{14.79 m g}{57.9121}; g \neq = 0

Simplificar

sin (2 θ_{0}) = 0.25538 \dots m g; g \neq = 0

sin (2 θ_{0}) = 0.25538 \dots m g; g \neq = 0

Aplique as propriedades trigonométricas inversas

sin (2 θ_{0}) = 0.25538 \dots m g

Soluções gerais para

sin (2 θ_{0}) = 0.25538 \dots m g

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

2 θ_{0} = arcsin (0.25538 \dots m g) + 2 πn, 2 θ_{0} = π + arcsin (- 0.25538 \dots m g) + 2 πn

2 θ_{0} = arcsin (0.25538 \dots m g) + 2 πn, 2 θ_{0} = π + arcsin (- 0.25538 \dots m g) + 2 πn

Resolver

2 θ_{0} = arcsin (0.25538 \dots m g) + 2 πn : θ_{0} = \frac{arcsin ( 0.25538 \dots m g )}{2} + πn

2 θ_{0} = arcsin (0.25538 \dots m g) + 2 πn

Dividir ambos os lados por

2

2 θ_{0} = arcsin (0.25538 \dots m g) + 2 πn

Dividir ambos os lados por

2

\frac{2 θ _{0}}{2} = \frac{arcsin ( 0.25538 \dots m g )}{2} + \frac{2 πn}{2}

Simplificar

θ_{0} = \frac{arcsin ( 0.25538 \dots m g )}{2} + πn

θ_{0} = \frac{arcsin ( 0.25538 \dots m g )}{2} + πn

Resolver

2 θ_{0} = π + arcsin (- 0.25538 \dots m g) + 2 πn : θ_{0} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + πn

2 θ_{0} = π + arcsin (- 0.25538 \dots m g) + 2 πn

Dividir ambos os lados por

2

2 θ_{0} = π + arcsin (- 0.25538 \dots m g) + 2 πn

Dividir ambos os lados por

2

\frac{2 θ _{0}}{2} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + \frac{2 πn}{2}

Simplificar

θ_{0} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + πn

θ_{0} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + πn

θ_{0} = \frac{arcsin ( 0.25538 \dots m g )}{2} + πn, θ_{0} = \frac{π}{2} + \frac{arcsin ( - 0.25538 \dots m g )}{2} + πn

2 cos (x) + 3 cos (x) - 2 = 0

sec (\frac{5 θ}{4}) = 2, [0.2 π]

cos (x) + ∣ cos (x) ∣ = 1, - 2 π \leq x \leq 2 π

\frac{30.56}{sin ( 8 8 ^{\circ} )} = \frac{17}{sin ( x )}

sec (x + \frac{π}{4}) = - 2, - π \leq x \leq π