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tan(-60)=((1-cos(u)))/(sin(u))

Solución

tan (- 6 0^{\circ}) = \frac{( 1 - cos ( u ) )}{sin ( u )}

u = 72 0^{\circ} n, u = 36 0^{\circ} + 72 0^{\circ} n, u = 24 0^{\circ} + 36 0^{\circ} n

Pasos de solución

tan (- 6 0^{\circ}) = \frac{( 1 - cos ( u ) )}{sin ( u )}

tan (- 6 0^{\circ}) = - 3

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

Restar

\frac{1 - cos ( u )}{sin ( u )}

de ambos lados

- 3 - \frac{1 - cos ( u )}{sin ( u )} = 0

Simplificar

- 3 - \frac{1 - cos ( u )}{sin ( u )} : \frac{- 3 sin ( u ) - 1 + cos ( u )}{sin ( u )}

\frac{- 3 sin ( u ) - 1 + cos ( u )}{sin ( u )} = 0

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

- 3 sin (u) - 1 + cos (u) = 0

Sea:

v = \frac{1}{2} u

- 3 sin (2 v) - 1 + cos (2 v) = 0

Re-escribir usando identidades trigonométricas

- 2 sin^{2} (v) - 2 cos (v) sin (v) 3 = 0

Factorizar

- 2 sin^{2} (v) - 2 cos (v) sin (v) 3 : - 2 sin (v) (sin (v) + 3 cos (v))

- 2 sin (v) (sin (v) + 3 cos (v)) = 0

Resolver cada parte por separado

sin (v) = 0 or sin (v) + 3 cos (v) = 0

sin (v) = 0 : v = 36 0^{\circ} n, v = 18 0^{\circ} + 36 0^{\circ} n

sin (v) + 3 cos (v) = 0 : v = 12 0^{\circ} + 18 0^{\circ} n

Combinar toda las soluciones

v = 36 0^{\circ} n, v = 18 0^{\circ} + 36 0^{\circ} n, v = 12 0^{\circ} + 18 0^{\circ} n

Sustituir en la ecuación

v = \frac{1}{2} u

\frac{1}{2} u = 36 0^{\circ} n : u = 72 0^{\circ} n

\frac{1}{2} u = 18 0^{\circ} + 36 0^{\circ} n : u = 36 0^{\circ} + 72 0^{\circ} n

\frac{1}{2} u = 12 0^{\circ} + 18 0^{\circ} n : u = 24 0^{\circ} + 36 0^{\circ} n

u = 72 0^{\circ} n, u = 36 0^{\circ} + 72 0^{\circ} n, u = 24 0^{\circ} + 36 0^{\circ} n

Siendo que la ecuación esta indefinida para:

72 0^{\circ} n, 36 0^{\circ} + 72 0^{\circ} n

u = 72 0^{\circ} n, u = 36 0^{\circ} + 72 0^{\circ} n, u = 24 0^{\circ} + 36 0^{\circ} n

sec^{2} (θ) (1 + cos^{2} (θ)) = 2 - tan^{2} (θ)

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

8 tan (x) - 12 = 5 tan (x) - 9

64 = arctan (\frac{x}{11})

2 sin (3 x) cos (3 x) = 1