Question #43388

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Answer 1 · 2014-11-12T19:52:50

#6sin^2x-cosx-4=0#

by #sin^2=1-cos^2x#,

#=>6(1-cos^2x)-cosx-4=0#

by cleaning up a bit,

#=>-6cos^2x-cosx+2=0#

by multiplying by #-1#,

#=> 6cos^2x+cosx-2=0#

by factoring out,

#=> (3cosx+2)(2cosx-1)=0#

by setting each factor equal to zero,

#=>{(cosx=-2/3 => x=cos^{-1}(2/3)", "-cos^{-1}(2/3)+2pi),(cosx=1/2 => x=pi/3","{5pi}/3):}#

#=>{(x=cos^{-1}(2/3)approx48.2^circ),(x=2pi-cos^{-1}(2/3) approx 311.8^circ),(x=pi/3=60^circ),(x={5pi}/3=300^circ):}#

I hope that this was helpful.