Question #0c379

2 Answers
Narad T.
Nov 13, 2017

The solutions are #S={1/4pi+2kpi, 3/4pi+2kpi,5/4pi+2kpi,7 /4pi+2kpi}#, #AA k inZZ#

Explanation:

We need

#csc theta=1/sintheta#

The equation is

#4sintheta-2csctheta=0#

#4sintheta-2/sintheta=0#

#2((2sin^2theta-1)/sintheta)=0#

Therefore,

#2sin^2theta-1=0#

#sin^2theta=1/2#

#sintheta=+-1/sqrt2#

#sintheta=1/sqrt2#, #=>#, #S={1/4pi+2kpi, 3/4pi+2kpi}#, #AA k in ZZ#

#sintheta=-1/sqrt2#, #=>#, #S={5/4pi+2kpi, 7/4pi+2kpi}#, #AA k in ZZ#

Somebody N.
Nov 13, 2017

#pi/4 , (3pi)/4#

Explanation:

Identity: #color(red)(cscx=1/sinx)#
..................................................................................................................................

#4sintheta-2csctheta=0#

#->4sintheta-2/sintheta=0#

Multiply by #sintheta#:

#4sin^2theta-2=0#

#4sin^2theta=2#

#sin^2theta=2/4=1/2#

#sintheta=1/sqrt(2)#

#theta=sin^-1(sintheta) =sin^-1(1/sqrt(2))=pi/4#

For interval: #0<=theta<=2pi#

#pi/4 , (3pi)/4#

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