2cos2xcosx+2cos3xcosx=0
⇒ 2cosx(cos2x+cos3x)=0
2cosx2cos25xcos2x=0
x=2π,23π,π,5π,53π,57π,59π
7Solutions.
If 0≤x<2π, then the number of real values of x, which satisfy the equation cosx+cos2x+cos3x+cos4x=0, is
Held on 3 Apr 2016 · Verified 6 Jul 2026.
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