How do you solve $2cosx + sin2x = 0$ in the interval [0, 2pi]?

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Answer 1 · 2016-07-05T13:13:59

$x= pi/2 or (3pi)/2$

$2cosx+sin2x=0$

$=>2cosx+2sinxcosx=0$

$=>2cosx(1+sinx)=0$

$=>cosx=0 and 1+sinx=0$

when
$cosx=0=cos(pi/2)=cos(2pi-pi/2)$

$:.x= pi/2 or x=(3pi)/2$

Again when

$1+sinx=0$

$=>sinx=-1=-sin(pi/2)=sin(pi+pi/2)$

$x=(3pi)/2$

How do you solve 2cosx + sin2x = 02cosx + sin2x = 0 in the interval [0, 2pi]?