How do you solve #2sin^2 theta-sin theta=0#?

2 Answers
Alan P.
Apr 18, 2016

#theta = n*pi# or
#theta= pi/6+n*2pi# or
#theta=(5pi)/6 +n*2pi#
#color(white)("XXXXXXXXXXXXXXXX")nin ZZ#

Explanation:

If #2sin^2theta-sin theta = 0#
then
#color(white)("XXX")(sin theta)(2sin theta -1)= 0#

#rarr color(white)("XXX")sin theta=0color(white)("XXXXX")orcolor(white)("XXXXX")2sin theta =1#
#color(white)("XXXX")rarr theta = 0+n*pi color(white)("XXXXXXXX")rarr theta=pi/6+2npi or (5pi)/6+2npi#

Gió
Apr 18, 2016

I tried to solve it as a normal quadratic equation in #sintheta# instead of #x#:

Explanation:

I would collect #sintheta# and write:

#sintheta[2sintheta-1]=0#

so:

#sintheta=0# when #theta=0,pi,2pi...# or #theta=npi# with #n=0,1,2,3,4....#

and:

#2sintheta-1=0#
#sintheta=1/2# when #theta=pi/6,5/6pi....# and multiples of #2npi# each.

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