√2 cos²x+cosx=0

Solve for x on interval [0,2π).
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1 Answer

sqrt(2)cos^2x+cosx=0; cosx(sqrt(2)cosx+1)=0, so cosx=0 (x=(pi)/2 and 3(pi)/2) and cosx=-sqrt(2)/2 (x=3(pi)/4 and 5(pi)/4). For angles in degrees use (pi)=180 degrees, giving x=90, 270, 135, 225.

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