If theta in left[-frac7pi6, frac4pi3right]$\theta \in \left[-\frac{7\pi}{6}, \frac{4\pi}{3}\right]$, then the number of solutions of sqrt3 csc^2 theta - 2left(sqrt3 - 1right) csc theta - 4 = 0$\sqrt{3} \csc^2 \theta - 2\left(\sqrt{3} - 1\right) \csc \theta - 4 = 0$, is equal to

### Related Formula sin(A pm B) = sin A cos B pm cos A sin B$$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$$ ### Core Logic Given equation: 3sin(alpha + beta) = 2sin(alpha - beta)$$3\sin(\alpha + \beta) = 2\sin(\alpha - \beta)$$ Expanding both sides: 3(sinalphacosbeta + cosalphasinbeta) = 2(sinalphacosbeta - cosalphasinbeta)$$3(\sin\alpha\cos\beta + \cos\alpha\sin\beta) = 2(\sin\alpha\cos\beta - \cos\alpha\sin\beta)$$ 3sinalphacosbeta + 3cosalphasinbeta = 2sinalphacosbeta - 2cosalphasinbeta$$3\sin\alpha\cos\beta + 3\cos\alpha\sin\beta = 2\sin\alpha\cos\beta - 2\cos\alpha\sin\beta$$ ### Step 1: Rearranging Terms Grouping like terms together: 3sinalphacosbeta - 2sinalphacosbeta = -2cosalphasinbeta - 3cosalphasinbeta$$3\sin\alpha\cos\beta - 2\sin\alpha\cos\beta = -2\cos\alpha\sin\beta - 3\cos\alpha\sin\beta$$ sinalphacosbeta = -5cosalphasinbeta$$\sin\alpha\cos\beta = -5\cos\alpha\sin\beta$$ Dividing both sides by cosalphacosbeta$\cos\alpha\cos\beta$: fracsinalphacosalpha = -5fracsinbetacosbeta$$\frac{\sin\alpha}{\cos\alpha} = -5\frac{\sin\beta}{\cos\beta}$$ tanalpha = -5tanbeta$$\tan\alpha = -5\tan\beta$$ ### Step 2: Conclusion Comparing with the given equation tanalpha = ktanbeta$\tan\alpha = k\tan\beta$, we get k = -5$k = -5$. *Note by our answer (Bonus)*: Since alpha, beta in (0, fracpi2)$\alpha, \beta \in (0, \frac{\pi}{2})$, both tanalpha$\tan\alpha$ and tanbeta$\tan\beta$ must be positive. Hence, tanalpha = -5tanbeta$\tan\alpha = -5\tan\beta$ is not possible. The data is inconsistent, but the NTA key marks option (2) as correct. ### Pattern Recognition Standard expansion of sin(Apm B)$\sin(A\pm B)$ and grouping identical products to isolate tan(A)$\tan(A)$ and tan(B)$\tan(B)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometric Functions

### 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

### Core Logic Let e^sin x = t$e^{\sin x} = t$, where t > 0$t > 0$ because exponential functions are strictly positive. Substitute into the equation: t - frac2t = 2$$t - \frac{2}{t} = 2$$ t^2 - 2t - 2 = 0$$t^2 - 2t - 2 = 0$$ Solve for t$t$ using the quadratic formula: t = frac2 pm sqrt4 - 4(1)(-2)2 = 1 pm sqrt3$$t = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = 1 \pm \sqrt{3}$$ Since t > 0$t > 0$, we discard 1 - sqrt3$1 - \sqrt{3}$. Thus, t = 1 + sqrt3 approx 2.732$t = 1 + \sqrt{3} \approx 2.732$. Now, equate back: e^sin x = 1 + sqrt3 implies sin x = ln(1 + sqrt3)$$e^{\sin x} = 1 + \sqrt{3} \implies \sin x = \ln(1 + \sqrt{3})$$ We know e approx 2.718$e \approx 2.718$. Since 1 + sqrt3 > e$1 + \sqrt{3} > e$, it follows that ln(1 + sqrt3) > 1$\ln(1 + \sqrt{3}) > 1$. But the range of sin x$\sin x$ is [-1, 1]$[-1, 1]$. Therefore, sin x$\sin x$ cannot equal a value strictly greater than 1. No real solution exists. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Trigonometric Functions Class 12 Maths: Continuity and Differentiability