### Related Formula sin 2x = 2 sin x cos x$\sin 2x = 2 \sin x \cos x$ ### Core Logic Given equation: 2 sin^3 x + sin 2x cos x + 4 sin x - 4 = 0$2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0$ Expand sin 2x$\sin 2x$: 2 sin^3 x + 2 sin x cos^2 x + 4 sin x - 4 = 0$2 \sin^3 x + 2 \sin x \cos^2 x + 4 \sin x - 4 = 0$ Factor out 2 sin x$2 \sin x$ from the first two terms: 2 sin x (sin^2 x + cos^2 x) + 4 sin x - 4 = 0$2 \sin x (\sin^2 x + \cos^2 x) + 4 \sin x - 4 = 0$ Since sin^2 x + cos^2 x = 1$\sin^2 x + \cos^2 x = 1$: 2 sin x (1) + 4 sin x - 4 = 0$2 \sin x (1) + 4 \sin x - 4 = 0$ 6 sin x - 4 = 0 Rightarrow sin x = frac46 = frac23$6 \sin x - 4 = 0 \Rightarrow \sin x = \frac{4}{6} = \frac{2}{3}$ ### Step 1: Finding appropriate interval for exactly 3 roots We need exactly 3 solutions in left[0, fracnpi2right]$\left[0, \frac{n\pi}{2}\right]$. The line y = 2/3$y = 2/3$ intersects the sine wave y = sin x$y = \sin x$ twice in every 2pi$2\pi$ interval. In [0, pi]$[0, \pi]$, there are 2 solutions. In [pi, 2pi]$[\pi, 2\pi]$, there are 0 solutions. In [2pi, 3pi]$[2\pi, 3\pi]$, there are 2 solutions (total 4 solutions). To get exactly 3 solutions, the interval must stretch past the first root in [2pi, 3pi]$[2\pi, 3\pi]$, but not reach the second root in that interval. However, the interval is defined as fracnpi2$\frac{n\pi}{2}$. Let's check endpoints fracnpi2$\frac{n\pi}{2}$: For n=4$n=4$: [0, 2pi]$[0, 2\pi]$ has 2 solutions. For n=5$n=5$: [0, frac5pi2]$[0, \frac{5\pi}{2}]$ includes [2pi, 2pi + fracpi2]$[2\pi, 2\pi + \frac{\pi}{2}]$. Since sin x = 2/3$\sin x = 2/3$ happens in (0, pi/2)$(0, \pi/2)$, there is exactly 1 solution in [2pi, 5pi/2]$[2\pi, 5\pi/2]$. Thus, total solutions = 3 for n=5$n=5$. ### Step 2: Solving quadratic equation Given n = 5$n = 5$, the quadratic equation is: x^2 + 5x + 2 = 0$x^2 + 5x + 2 = 0$ Using quadratic formula: x = frac-5 pm sqrt25 - 82 = frac-5 pm sqrt172$x = \frac{-5 \pm \sqrt{25 - 8}}{2} = \frac{-5 \pm \sqrt{17}}{2}$ The roots are approximately frac-5 pm 4.122$\frac{-5 \pm 4.12}{2}$, which evaluates to roughly -0.44$-0.44$ and -4.56$-4.56$. Both roots are strictly negative. ### Step 3: Determining interval membership Since both roots are negative, they belong to the interval (-infty, 0)$(-\infty, 0)$. ### Pattern Recognition Collapsing complex trigonometric expressions often yields c_1sin x = c_2$c_1\sin x = c_2$. Overlaying horizontal line intersections on the sine graph bounds n$n$ rapidly by counting nodes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometric Functions Class 11 Maths: Complex Numbers and Quadratic Equations