Question #cf198

1 Answer
Jim G.
Nov 28, 2017

#"see explanation"#

Explanation:

#"using the "color(blue)"trigonometric identity"#

#•color(white)(x)cos2x=2cos^2x-1#

#rArr2(2cos^2x-1)-5cosx-4=0#

#rArr4cos^2x-5cosx-6=0larrcolor(blue)"quadratic in cos"#

#rArr4cos^2x-8cosx+3cosx-6=0#

#rArr4cosx(cosx-2)+3(cosx-2)=0#

#rarr(cosx-2)(4cosx+3)=0#

#"equate each factor to zero and solve for x"#

#cosx-2=0rArrcosx=2larrcolor(red)"no solution"#

#4cosx+3=0rArrcosx=-3/4#

#cosx<0" hence x is in second/third quadrants"#

#rArrx=cos^-1(3/4)=0.723larrcolor(blue)"related acute angle"#

#rArrx=(pi-0.723)=2.419larrcolor(blue)"second quadrant"#

#rArrx=(pi+0.723)=3.865larrcolor(blue)"third quadrant"#

#"in general the solutions are"#

#x=2.419+2kpi" or "x=3.865+2kpi;k inZZ#

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