Question #b8716

2 Answers
SCooke
Aug 10, 2017

Removed previous post.

Explanation:

Apparently incorrect. Sorry.

Monzur R.
Aug 10, 2017

#sinx = npi, n in ZZ#

#sinx = 1/2pi+2npi, n in ZZ#

Explanation:

#sin^2x=sinx#

#sin^2x-sinx=0#

#sinx(sinx-1)=0#

#sinx=0 rArr x= npi, n in ZZ#

#sinx-1=0#

#sinx=1rArrx=1/2pi+2npi, n in ZZ#

Note: we cannot just divide through by #sinx# and then solve #sinx=1#. This is because #sinx# can equal zero and division by zero is undefined. So what we need to do instead is get the equation in standard form, factor out any common factors and then solve for #x#.

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