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 $x=5/3pi+2kpi$ .

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