solve sin 2x-cos x=0 for -pi=<x=<2pi

using double angle formula and factoring
in Trigonometry Answers by Level 1 User (400 points)

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1 Answer

sin(2x)-cos(x)=0,

2sin(x)cos(x)-cos(x)=0=cos(x)(2sin(x)-1), therefore cos(x)=0, x=π/2, 3π/2 or -π/2; or sin(x)=½, x=π/6 or 5π/6.

by Top Rated User (1.2m points)

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