How do you solve $(tanx + 1)(sinx - 1) = 0$ ?

1 Answer

Dec 6, 2016

$x=-pi/4$ , $x=pi/2$

Explanation:

$(tanx+1)(sinx-1)=0$
It means
either $tanx+1=0$ or $sinx-1=0$
Therefore we can consider both situations i.e.
$tanx=-1$
$x=tan^-1-1$
$x=-pi/4$
similarly
$sinx-1=0$
$sinx=1$
$x=sin^-1 1$
$x=pi/2$

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How do you solve $(tanx + 1)(sinx - 1) = 0$ ?

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