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#

Answer link

Impact of this question

7338 views around the world

You can reuse this answer
Creative Commons License

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question