Further Pure Mathematics 4PM1 / Trigonometry / 1.161.16: Solve trigonometric equations4PM1/1R/November/2020 — Question 10 · 11 marksMark as done · Save for laterSolve(a) sin(x+π3)=32,0≤x≤2π,\mathrm{sin}\left(x+\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}, \qquad 0\leq x\leq2\pi,sin(x+3π)=23,0≤x≤2π,giving your answers in terms of π\displaystyle \piπ,(3)(b) 3sin θ+5cos θ=0,−360∘≤θ≤360∘,3\mathrm{sin}\,\theta+5\mathrm{cos}\,\theta=0, \qquad -360^\circ\leq\theta\leq360^\circ,3sinθ+5cosθ=0,−360∘≤θ≤360∘,giving your answers to the nearest degree,(3)(c) 1+sin 2y=2cos22y,−180∘≤y≤0∘.1+\mathrm{sin}\,2y=2\mathrm{cos}^2 2y, \qquad -180^\circ\leq y\leq0^\circ.1+sin2y=2cos22y,−180∘≤y≤0∘.(5)▸ Mark scheme← 1.15: 4PM1/1R/November/2020 — Question 61.17: 4PM1/2/November/2020 — Question 10 →
