Question

calcula x donde :tan(3pi/2-5x)=cot (x-pi/9)

Answer 1

Respuesta:

$tan\left(\frac{3\pi }{2}-5x\right)=cot\:\left(x-\frac{\pi }{9}\right)$

$\frac{\cos \left(5x\right)}{\sin \left(5x\right)}=\cot \left(x-\frac{\pi }{9}\right)$

Restamos cot (x - π/9) de ambos lados

$\cot \left(x-\frac{\pi }{9}\right)$

$\boxed{\frac{\cos \left(5x\right)}{\sin \left(5x\right)}-\cot \left(x-\frac{\pi }{9}\right)=0}$

Simplificamos

$\frac{\cos \left(5x\right)}{\sin \left(5x\right)}-\cot \left(x-\frac{\pi }{9}\right)=\frac{\cos \left(5x\right)-\cot \left(\frac{9x-\pi }{9}\right)\sin \left(5x\right)}{\sin \left(5x\right)}$

$\boxed{\frac{\cos \left(5x\right)-\cot \left(\frac{9x-\pi }{9}\right)\sin \left(5x\right)}{\sin \left(5x\right)}=0}$

$\frac{f\left(x\right)}{g\left(x\right)}=0\quad \Rightarrow \quad f\left(x\right)=0$

$\cos \left(5x\right)-\cot \left(\frac{9x-\pi }{9}\right)\sin \left(5x\right)=0$

Expresar con seno, coseno

$\frac{\cos \left(5x\right)\sin \left(\frac{-\pi +9x}{9}\right)-\cos \left(\frac{-\pi +9x}{9}\right)\sin \left(5x\right)}{\sin \left(\frac{-\pi +9x}{9}\right)}=0$

$\cos \left(5x\right)\sin \left(\frac{-\pi +9x}{9}\right)-\cos \left(\frac{-\pi +9x}{9}\right)\sin \left(5x\right)=0$

$-\sin \left(5x-\frac{-\pi +9x}{9}\right)=0$

Dividimos entre -1 ambos lados

$\frac{-\sin \left(5x-\frac{-\pi +9x}{9}\right)}{-1}=\frac{0}{-1}$

$\boxed{\sin \left(5x-\frac{-\pi +9x}{9}\right)=0}$

$5x-\frac{-\pi +9x}{9}=0+2\pi$

$5x-\frac{-\pi +9x}{9}=\pi +2\pi$

Resolvemos

$5x-\frac{-\pi +9x}{9}=0+2\pi$        ⇒       $x=\frac{\pi n}{2}-\frac{\pi }{36}$

$5x-\frac{-\pi +9x}{9}=\pi +2\pi$        ⇒       $\:x=\frac{2\pi }{9}+\frac{\pi n}{2}$

Soluciones

$\boxed{x=\frac{\pi n}{2}-\frac{\pi }{36}}$

$\boxed{x=\frac{2\pi }{9}+\frac{\pi n}{2}}$

Saludos...