How do you find the exact value of #4sin^2theta-3sintheta=1# in the interval #0<=theta<2pi#?

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Answer 1 · 2016-12-23T18:17:16

#theta={pi/2,3.39,6.03}#

#4sin^2theta-3sintheta=1#

#hArr4sin^2theta-3sintheta-1=0#

or #4sin^2theta-4sintheta+sintheta-1=0#

or #4sintheta(sintheta-1)+1(sintheta-1)=0#

or #(4sintheta+1)(sintheta-1)=0#

therefore either #4sintheta+1=0# i.e. #sintheta=-1/4# and #theta=345.52^@=6.03# or #194.48^@=3.39#

or #sintheta=1# i.e. #theta=pi/2#

Hence #theta={pi/2,3.39,6.03}#