# Solve limit (as x approaches 0) of 1

I am trying to lớn solve this equation , but i get different answers from the book. Can someone help me please?How to solve

\$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 \$  Thanks for contributing an answer lớn ccevents.vnematics Stack Exchange!

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