How do I find values of Θ between 0 and 2π that satisfy -1 - 3 sin Θ = cos 2Θ ??

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Answer 1 · 2018-02-22T19:33:24

#(7pi)/6color(white)(888)# III Quadrant

#(11pi)/6color(white)(88)# IV Quadrant

#-1-3sintheta=cos(2theta)#

Identity:

#color(red)bb(cos2x=1-2sin^2x)#

#-1-3sintheta=1-2sin^2x#

Subtract #(1-2sin^2theta)# from both sides:

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

Simplify and rearrange:

#2sin^2theta-3sintheta-2=0#

Let #u=sintheta#

#:.#

#2u^2-3u-2=0#

Factor:

#(2u+1)(u-2)=0=>u=-1/2 and u=2#

But #u=sintheta#

#:.#

#sintheta=-1/2 and sintheta=2#

#sintheta=2# has now real solution, because:

#-1<=sintheta<=1#

#sintheta=-1/2#

#theta=arcsin(sintheta)=arcsin(-1/2)=>theta=-pi/6+2pi#

#=(11pi)/6 and pi/6+pi=(7pi)/6#

So we have:

#(7pi)/6color(white)(888)# III Quadrant

#(11pi)/6color(white)(88)# IV Quadrant