How do you solve algebraically for x in #cos2x+sinx=0#?

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Answer 1 · 2015-02-15T20:47:09

The answers are: #x=pi/2+2kpi#, #x=4/3pi# and #x=5/3pi+2kpi#.

#cos2x+sinx=0rArr1-2sin^2x+sinx=0rArr#

#2sin^2x-sinx-1=0#

This is an equation of second degree, so we can apply the resolution formula:

#Delta=b^2-4ac=1-4*2*(-1)=9#,

#sinx=(-b+-sqrtDelta)/(2a)=(1+-3)/(2*2)rArr#

#(sinx)_1=1# and
#(sinx)_2=-1/2#.

The first gives: #x=pi/2+2kpi#

The second gives: #x=4/3pi# and #x=5/3pi+2kpi#.