Solve limit (as x approaches 0) of 1
$cos(x)-cos(2x)+cos(3x)=0$
My answers is : $x=45^circ+90^circ k$, $x=pmfrac12 +360^circ k$
Book"s answer: $pm60^circ+360^circ k$ , $45^circ+90^circ k$
Notice, $$cos x-cos 2x+cos 3x=0$$ $$(cos x+cos 3x)-cos 2x=0$$$$2cosleft(fracx+3x2 ight)cosleft(fracx-3x2 ight)-cos 2x=0$$
$$2cos 2xcos x-cos 2x=0$$$$cos 2x(2cos x-1)=0 $$Now, solving for $x$ as follows $$cos 2x=0implies 2x=(2k+1)fracpi2$$$$x=(2k+1)fracpi4$$$$ colorredx=90^circ k+45^circ$$or $$2cos x-1=0$$$$ cos x=frac12=cos fracpi3$$$$x=2kpipmfracpi3$$$$colorredx=360^circ kpm 60^circ$$Where, $colorbluek$ is any integer
First we need to simplify $cos(x)-cos(2x)+cos(3x)=0$.
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$1-2cos(x)-2cos^2(x)+4cos^3(x)=0$
$(2cos(x)-1)(2cos^2(x)-1)=0$
Split lớn two equations:
$2cos(x)-1=0$ or $2cos^2(x)-1=0$
And it"s easy lớn solve:
$x=fracpi3+2pi C_1$ ;
$x=frac5 pi 3+2pi C_2$ ;
$x=fracpi4+2 pi C_3$ ;
$x=frac7pi4+2 pi C_4$.
$C_1,C_2,C_3,C_4 in ccevents.vnbbZ $
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