How do you find the exact value of $2sin2theta+costheta=0$ in the interval $0<=theta<2pi$ ?

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Answer 1 · 2018-04-11T09:35:08

$color(blue)(pi/2,(3pi)/2,2pi-arcsin(1/4),pi+arcsin(1/4))$

Identity:

$color(red)bb(sin(2x)=2sinxcosx)$

Substituting this in given equation:

$2(2sin(theta)cos(theta)+cos(theta)=0$

$4sin(theta)cos(theta)+cos(theta)=0$

Factoring out $cos(theta)$ :

$cos(theta)[4sin(theta)+1]=0$

$cos(theta)=0$

$theta=arccos(cos(theta))=arccos(0)=> theta=pi/2,(3pi)/2$

$4sin(theta)+1=0$

$sin(theta)=-1/4$

$theta=arcsin(sin(theta))=arcsin(-1/4)$

This is in the IV quadrant. so given as a positive angle:

$2pi-arcsin(1/4)$

We also have an angle in the III quadrant, since the sine is negative:

$pi+arcsin(1/4)$

So our solutions are:

$color(blue)(pi/2,(3pi)/2,2pi-arcsin(1/4),pi+arcsin(1/4))$

How do you find the exact value of 2sin2theta+costheta=02sin2theta+costheta=0 in the interval 0<=theta<2pi0<=theta<2pi?