Additional Mathematics 0606 / Trigonometry / 2.232.23: Modulus sine, tan and cot equations0606/23/M/J/17 — Question 10 · 12 marksMark as done · Save for later · Show all solutions(a)Solve 2∣sinx∣=12|\sin x| = 12∣sinx∣=1 for −π≤x≤π-\pi \le x \le \pi−π≤x≤π radians.[3]▸ Answer(b)Solve 3tan(2y+15°)=13\tan(2y + 15^{\degree}) = 13tan(2y+15°)=1 for 0°≤y≤180°0^{\degree} \le y \le 180^{\degree}0°≤y≤180°.[4]▸ Answer(c)Solve 3cot2z=csc2z−7cscz+13\cot^{2} z = \csc^{2} z - 7\csc z + 13cot2z=csc2z−7cscz+1 for 0°≤z≤360°0^{\degree} \le z \le 360^{\degree}0°≤z≤360°.[5]▸ Answer▸ Official mark scheme← 2.14: Sine and secant quadratic equations1.28: Sketch of a modulus cosine graph →
