Respuesta :

We are to solve

[tex]5\tan ^2\beta+5\tan \beta=0[/tex]

Dividing through by 5 we get

[tex]\tan ^2\beta+\tan \beta=0[/tex]

Simplifying further we have

[tex]\tan \beta(1+\tan \beta)=0[/tex]

This gives

[tex]\tan \beta=0,\tan \beta+1=0[/tex]

For

[tex]\tan \beta=0[/tex]

Since for every multiple of π tan β = 0

Hence,

[tex]\beta=n\pi[/tex]

For

[tex]\tan \beta+1=0[/tex]

This gives

[tex]undefined[/tex]

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