Additional Mathematics 0606 / Trigonometry / 1.11.1: Factorising a trigonometric equation0606/11/M/J/16 — Question 9 · 7 marksMark as done · Save for later · Show all solutions(i)Show that 2cosxcotx+1=cotx+2cosx2\cos x \cot x + 1 = \cot x + 2\cos x2cosxcotx+1=cotx+2cosx can be written in the form (acosx−b)(cosx−sinx)=0(a\cos x - b)(\cos x - \sin x) = 0(acosx−b)(cosx−sinx)=0, where aaa and bbb are constants to be found.[4]▸ Answer(ii)Hence, or otherwise, solve 2cosxcotx+1=cotx+2cosx2\cos x \cot x + 1 = \cot x + 2\cos x2cosxcotx+1=cotx+2cosx for 0<x<π0 < x < \pi0<x<π.[3]▸ Answer▸ Official mark scheme← 2.28: Cosine rule with surds and area1.10: Secant identity and solving in radians →
