Find all values of x in the interval [0, 2π] that satisfy the equation. (Enter your answers as a comma-separated list.)? 4 + 2 cos(2x) = 6 cos(x)

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Answer 1 · 2018-01-16T07:23:54

# x in {0,pi/3,5pi/3} sub [0,2pi]#.

Recall that, #cos2x=2cos^2x-1#.

Hence, the eqn. becomes, #4+2(2cos^2x-1)=6cosx, i.e., #

#4cos^2x-6cosx+2=0, or, 2cos^2x-3cosx+1=0#.

#:. ul(2cos^2x-2cosx)-ul(cosx+1)=0#.

#:. 2cosx(cosx-1)-1(cosx-1)=0#.

#:. (cosx-1)(2cosx-1)=0#.

#:. cosx=1=cos0, or, cosx=1/2=cos(pi/3)#.

#:. x=0, or, x=pi/3.#

Since, #cos(2pi-pi/3)=cos(pi/3)=1/2, x=2pi-pi/3=5pi/3#.

#:. x in {0,pi/3,5pi/3} sub [0,2pi]#.