sin(2x)tan(2x+2)=sin(3x-1)tan(2x+2)

Solución

sin (2 x) tan (2 x + 2) = sin (3 x - 1) tan (2 x + 2)

x = \frac{πn}{2} - 1, x = \frac{π}{5} + \frac{4 πn}{5} + \frac{1}{5}, x = \frac{3 π}{5} + \frac{4 πn}{5} + \frac{1}{5}, x = - 4 πn + 1, x = 1 - 2 π - 4 πn

Pasos de solución

sin (2 x) tan (2 x + 2) = sin (3 x - 1) tan (2 x + 2)

Restar

sin (3 x - 1) tan (2 x + 2)

de ambos lados

sin (2 x) tan (2 x + 2) - sin (3 x - 1) tan (2 x + 2) = 0

Factorizar

sin (2 x) tan (2 x + 2) - sin (3 x - 1) tan (2 x + 2) : tan (2 (x + 1)) (sin (2 x) - sin (- 1 + 3 x))

tan (2 (x + 1)) (sin (2 x) - sin (- 1 + 3 x)) = 0

Resolver cada parte por separado

tan (2 (x + 1)) = 0 or sin (2 x) - sin (- 1 + 3 x) = 0

tan (2 (x + 1)) = 0 : x = \frac{πn}{2} - 1

sin (2 x) - sin (- 1 + 3 x) = 0 : x = \frac{π}{5} + \frac{4 πn}{5} + \frac{1}{5}, x = \frac{3 π}{5} + \frac{4 πn}{5} + \frac{1}{5}, x = - 4 πn + 1, x = 1 - 2 π - 4 πn

Combinar toda las soluciones

x = \frac{πn}{2} - 1, x = \frac{π}{5} + \frac{4 πn}{5} + \frac{1}{5}, x = \frac{3 π}{5} + \frac{4 πn}{5} + \frac{1}{5}, x = - 4 πn + 1, x = 1 - 2 π - 4 πn

2 sin^{2} (θ) - 7 sin (θ) - 4 = 0

2 tan (x) - 23 = 0

sin^{2} (x) + cos (x) = - 1

2 sin^{2} (\frac{x}{4}) - 3 cos (\frac{x}{4}) = 0

cos^{2} (A) + sin^{2} (A) = \frac{1}{sin ^{2} ( A )}