Respuesta :
Answer:[tex]x=\frac{\pi}{4}[/tex]
Step-by-step explanation:
`Writing tan and cos in form of sin
[tex]\frac{sinx}{cosx}-\left ( 1-2sin^2x\right )=2sinxcosx[/tex]
[tex]\frac{sinx}{cosx}-1+2sin^2x=2sinxcosx[/tex]
[tex]\frac{sinx-cosx}{cosx}=2sinx\left ( cosx-sinx\right )[/tex]
[tex]2sinx\left ( cosx-sinx\right )-\frac{sinx-cosx}{cosx}=0[/tex]
[tex]\left ( cosx-sinx\right )\left [ 2sinx+\frac{1}{cosx}\right ]=0[/tex]
[tex]\left ( cosx-sinx\right )\left [ \frac{2sinxcosx+1}{cosx}\right ]=0[/tex]
so cosx-sinx=0 or 2sinxcosx+1=0 and cosx is not equals to zero
cosx-sinx=0
tanx=1,[tex]x=\frac{\pi}{4}[/tex]
similarly
[tex]x=n\pi +\frac{\pi}{4}[/tex]
For 2sinxcosx+1=0
sin2x=-1
[tex]2x=\frac{3\pi}{2}[/tex]
but this value is not satisfied by equation thus
[tex]x=n\pi +\frac{\pi}{4}[/tex]
