How do you solve #3cos2theta=2-sintheta# in the interval #0<=x<=2pi#?

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Answer 1 · 2016-07-25T03:45:15

#theta=30^o,150^o,199.47^o# or #340.53^o#

#3cos2theta=2-sintheta# can be simplified as

#3(1-2sin^2theta)=2-sintheta# or

#3-6sin^2theta=2-sintheta#

#6sin^2theta-sintheta-1=0#

Hence #sintheta=(1+-sqrt(1^2-4xx6xx(-1)))/(2xx6)#

= #(1+-sqrt(1+24))/12=(1+-5)/12#

Hence either #sintheta=1/2# i.e. #theta=pi/6=30^o# or #(5pi)/6=150^o#

or #sintheta=-1/3#

now as #sin19.47^o=1/3#,

#theta=180^o+19.47^o=199.47^o# or #theta=360^o-19.47^o=340.53^o#