Solve the following equations for all solutions of x cotxcos^2x=2cotx
equate to zero .. . . (cot x)(cos^2 x) - (2cot x) = 0 then factor completely (cot x) (cos^2 x - 2) = 0 (cot x) (cos x - √2) (cos x + √2) = 0 then either cot x = 0 cos x = √2 ≈ 1.4 cos x = -√2 but cosine values cannot be outside -1 and 1 thus only cot x = 0 would yield solution and the answers are .. . .. . π/2 and 3π/2
is that helps u @KBeard
Yes, that helped so much. I am having trouble identifying where to start. Thank you!