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))#