Respuestas
Respuesta:
a)x=\frac{3\pi }{4}+\pi n,\:x=\frac{\pi }{4}+\pi nx=
4
3π
+πn,x=
4
π
+πn
Explicación paso a paso:
\frac{\tan (x)}{\cot (x)}-1=0
cot(x)
tan(x)
−1=0
\frac{\tan (x)-\cot (x)}{\cot (x)}=0
cot(x)
tan(x)−cot(x)
=0
\frac{f(x)}{g(x)}=0\quad \Rightarrow \quad f(x)=0
g(x)
f(x)
=0⇒f(x)=0
\tan (x)-\cot (x)=0tan(x)−cot(x)=0
-\frac{\cos (x)}{\sin (x)}+\frac{\sin (x)}{\cos (x)}=0−
sin(x)
cos(x)
+
cos(x)
sin(x)
=0
\frac{-\cos ^2(x)+\sin ^2(x)}{\cos (x)\sin (x)}=0
cos(x)sin(x)
−cos
2
(x)+sin
2
(x)
=0
-\cos ^2(x)+\sin ^2(x)=0−cos
2
(x)+sin
2
(x)=0
(\sin (x)+\cos (x))(\sin (x)-\cos (x))=0(sin(x)+cos(x))(sin(x)−cos(x))=0
\sin (x)+\cos (x)=0\quad \mathrm{o}\quad \sin (x)-\cos (x)=0sin(x)+cos(x)=0osin(x)−cos(x)=0
x=\frac{3\pi }{4}+\pi n,\:x=\frac{\pi }{4}+\pi nx=
4
3π
+πn,x=
4
π
+πn
b)\frac{senx+cosx}{senx}=1
senx
senx+cosx
=1
\frac{\sin (x)+\cos (x)}{\sin (x)}-1=0
sin(x)
sin(x)+cos(x)
−1=0
\frac{\cos (x)}{\sin (x)}=0
sin(x)
cos(x)
=0
\frac{f(x)}{g(x)}=0\quad \Rightarrow \quad f(x)=0
g(x)
f(x)
=0⇒f(x)=0
\cos (x)=0cos(x)=0
x=\frac{\pi }{2}+2\pi n,\:x=\frac{3\pi }{2}+2\pi nx=
2
π
+2πn,x=
2
3π
+2πn