We have cosx+cos‌2‌x+cos‌3‌x=0 or (cos‌3‌x+cosx)+cos‌2‌x=0 or 2‌cos‌2‌x⋅cosx+cos‌2‌x=0 or cos‌2‌x(2‌cosx+1)=0 We have either cos‌2‌x=0‌or‌2‌cosx+1=0 If
cos‌2‌x=0,‌then‌2x=(2m+1)‌‌
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2
or x=(2m+1)‌‌
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,m∈I If 2‌cosx+1=0,then cosx=−‌‌
1
2
=cos‌‌
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⇒x=2nπ±‌‌
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,n∈I Hence the required general solution are x=(2m+1)‌‌