(a)[3]
Write $5\sin\theta + 12\cos\theta$ as $R\cos(\theta - \alpha)$, with $R > 0$ and $0 < \alpha < \frac{1}{2}\pi$.
(b)[4]
Hence solve for $x$ in the equation $5\sin 2x + 12\cos 2x = 6$ for $0 \leq x \leq \pi$.
Mathematics 9709 · AS & A Level · Trigonometry
Write $5\sin\theta + 12\cos\theta$ as $R\cos(\theta - \alpha)$, with $R > 0$ and $0 < \alpha < \frac{1}{2}\pi$.
Hence solve for $x$ in the equation $5\sin 2x + 12\cos 2x = 6$ for $0 \leq x \leq \pi$.