How do you find the exact value of #5sin^2theta-4sintheta+cos2theta=0# in the interval #0<=theta<2pi#?

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Answer 1 · 2017-04-06T13:06:00

#theta={0.334,pi/2,(pi-0.334)}#

#5sin^2theta-4sintheta+cos2theta=0#

#hArr5sin^2theta-4sintheta+1-2sin^2theta=0#

or #3sin^2theta-4sintheta+1=0#

and #sintheta=(-(-4)+-sqrt(4^2-4xx3xx1))/(2xx3)#

= #(4+-2)/6# i.e. #sintheta=1=sin(pi/2)#

or #sintheta=1/3=sin0.334#

and #theta=2npi+pi/2# or #npi+(-1)^n0.334#

and in #0 <= theta <= 2pi#

#theta={0.334,pi/2,(pi-0.334)}#