Additional Mathematics 0606 / Trigonometry / 2.272.27: Sec–tan quadratic and sine equation0606/23/M/J/20 — Question 10 · 8 marksMark as done · Save for later · Show all solutions(a)Solve 5sec2A+14tanA−8=05\sec^{2} A + 14\tan A - 8 = 05sec2A+14tanA−8=0 for 0°≤A≤180°0^{\degree} \le A \le 180^{\degree}0°≤A≤180°.[4]▸ Answer(b)Solve 5sin(4B−π8)+2=05\sin\left(4B - \tfrac{\pi}{8}\right) + 2 = 05sin(4B−8π)+2=0 for −π4≤B≤π4-\tfrac{\pi}{4} \le B \le \tfrac{\pi}{4}−4π≤B≤4π radians.[4]▸ Answer▸ Official mark scheme← 2.26: Sine curve constants and tangent sketch1.20: Cosine graph sketch, amplitude and period →
