### Related Formula textQuadratic formula: y = frac-b pm sqrtb^2 - 4ac2a$\text{Quadratic formula: } y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ ### Core Logic This is a quadratic equation in terms of csc theta$\csc \theta$. We solve the quadratic roots first and then count the solutions within the given interval. ### Step 1: Solve the quadratic equation Let y = csc theta$y = \csc \theta$: sqrt3 y^2 - 2(sqrt3 - 1)y - 4 = 0$\sqrt{3} y^2 - 2(\sqrt{3} - 1)y - 4 = 0$ y = frac2(sqrt3-1) pm sqrt4(sqrt3-1)^2 - 4(sqrt3)(-4)2sqrt3$y = \frac{2(\sqrt{3}-1) \pm \sqrt{4(\sqrt{3}-1)^2 - 4(\sqrt{3})(-4)}}{2\sqrt{3}}$ y = frac2(sqrt3-1) pm sqrt4(3 + 1 - 2sqrt3) + 16sqrt32sqrt3$y = \frac{2(\sqrt{3}-1) \pm \sqrt{4(3 + 1 - 2\sqrt{3}) + 16\sqrt{3}}}{2\sqrt{3}}$ y = frac2(sqrt3-1) pm sqrt16 + 8sqrt32sqrt3$y = \frac{2(\sqrt{3}-1) \pm \sqrt{16 + 8\sqrt{3}}}{2\sqrt{3}}$ Since 16 + 8sqrt3 = (2 + 2sqrt3)^2$16 + 8\sqrt{3} = (2 + 2\sqrt{3})^2$: y = frac2(sqrt3-1) pm (2 + 2sqrt3)2sqrt3$y = \frac{2(\sqrt{3}-1) \pm (2 + 2\sqrt{3})}{2\sqrt{3}}$ - Case 1 (+$+$ sign): y = frac4sqrt32sqrt3 = 2 implies sin theta = frac12$y = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 \implies \sin \theta = \frac{1}{2}$ - Case 2 (-$-$ sign): y = frac-42sqrt3 = -frac2sqrt3 implies sin theta = -fracsqrt32$y = \frac{-4}{2\sqrt{3}} = -\frac{2}{\sqrt{3}} \implies \sin \theta = -\frac{\sqrt{3}}{2}$ ### Step 2: Count solutions in the interval Our interval is theta in left[-frac7pi6, frac4pi3 ight]$\theta \in \left[-\frac{7\pi}{6}, \frac{4\pi}{3} ight]$: - For sin theta = frac12$\sin \theta = \frac{1}{2}$: The general solutions are theta = fracpi6, frac5pi6$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$. Within our interval, we have: theta = -frac7pi6, \, fracpi6, \, frac5pi6 quad (3 text solutions)$\theta = -\frac{7\pi}{6}, \, \frac{\pi}{6}, \, \frac{5\pi}{6} \quad (3 \text{ solutions})$ - For sin theta = -fracsqrt32$\sin \theta = -\frac{\sqrt{3}}{2}$: The general solutions are theta = -fracpi3, -frac2pi3$\theta = -\frac{\pi}{3}, -\frac{2\pi}{3}$. Within our interval, we have: theta = -frac2pi3, \, -fracpi3, \, frac4pi3 quad (3 text solutions)$\theta = -\frac{2\pi}{3}, \, -\frac{\pi}{3}, \, \frac{4\pi}{3} \quad (3 \text{ solutions})$ Summing the valid solutions: textTotal solutions = 3 + 3 = 6$\text{Total solutions} = 3 + 3 = 6$ ### Pattern Recognition Perfect Square discriminant: In JEE quadratics with irrational coefficients, the discriminant b^2-4ac$b^2-4ac$ almost always simplifies to a perfect square of the form (p + qsqrtr)^2$(p + q\sqrt{r})^2$. Double check your algebraic expansions if it doesn't simplify cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions

### Related Formula Factorization of quadratic equations by splitting the middle term. ### Core Logic Treat the given equation as a standard quadratic in terms of costheta$\cos\theta$ and solve for its roots. ### Step 1: Factorization Split the middle term: 2sqrt2cos^2theta + 2costheta - sqrt6costheta - sqrt3 = 0$2\sqrt{2}\cos^2\theta + 2\cos\theta - \sqrt{6}\cos\theta - \sqrt{3} = 0$ 2costheta(sqrt2costheta + 1) - sqrt3(sqrt2costheta + 1) = 0$2\cos\theta(\sqrt{2}\cos\theta + 1) - \sqrt{3}(\sqrt{2}\cos\theta + 1) = 0$ (2costheta - sqrt3)(sqrt2costheta + 1) = 0$(2\cos\theta - \sqrt{3})(\sqrt{2}\cos\theta + 1) = 0$ ### Step 2: Solve for roots This yields two possible cases: costheta = fracsqrt32 quad textor quad costheta = -frac1sqrt2$\cos\theta = \frac{\sqrt{3}}{2} \quad \text{or} \quad \cos\theta = -\frac{1}{\sqrt{2}}$ ### Step 3: Count Solutions in Interval The given interval is [-2pi, 2pi]$[-2\pi, 2\pi]$, which covers two complete cycles of the cosine wave. * For costheta = fracsqrt32$\cos\theta = \frac{\sqrt{3}}{2}$, there are 2$2$ solutions per cycle implies 2 times 2 = 4$\implies 2 \times 2 = 4$ solutions. * For costheta = -frac1sqrt2$\cos\theta = -\frac{1}{\sqrt{2}}$, there are 2$2$ solutions per cycle implies 2 times 2 = 4$\implies 2 \times 2 = 4$ solutions. textTotal Solutions = 4 + 4 = 8$\text{Total Solutions} = 4 + 4 = 8$ ### Pattern Recognition Since both fracsqrt32$\frac{\sqrt{3}}{2}$ and -frac1sqrt2$-\frac{1}{\sqrt{2}}$ lie strictly between -1$-1$ and 1$1$, each horizontal line cuts the cosine function exactly twice per period (2pi$2\pi$). Across an interval of width 4pi$4\pi$, each value must yield exactly 4$4$ solutions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions

### Related Formula We must reduce the equation to a single trigonometric ratio (like sin x$\sin x$) using: cos^2 x = 1 - sin^2 x$\cos^2 x = 1 - \sin^2 x$ Rationalizing fractions: frac1A + sqrtB = fracA - sqrtBA^2 - B$\frac{1}{A + \sqrt{B}} = \frac{A - \sqrt{B}}{A^2 - B}$ ### Core Logic Simplify the constant term on the RHS: -frac41 + sqrt3 = -frac4(sqrt3 - 1)2 = -2(sqrt3 - 1) = 2 - 2sqrt3$-\frac{4}{1 + \sqrt{3}} = -\frac{4(\sqrt{3} - 1)}{2} = -2(\sqrt{3} - 1) = 2 - 2\sqrt{3}$ Now replace cos^2 x = 1 - sin^2 x$\cos^2 x = 1 - \sin^2 x$ in the main equation: (4 - sqrt3)sin x - 2sqrt3(1 - sin^2 x) = 2 - 2sqrt3$(4 - \sqrt{3})\sin x - 2\sqrt{3}(1 - \sin^2 x) = 2 - 2\sqrt{3}$ 2sqrt3sin^2 x + (4 - sqrt3)sin x - 2 = 0$2\sqrt{3}\sin^2 x + (4 - \sqrt{3})\sin x - 2 = 0$ ### Step 1: Solving the quadratic in sin x$\sin x$ Let y = sin x$y = \sin x$: 2sqrt3y^2 + 4y - sqrt3y - 2 = 0$2\sqrt{3}y^2 + 4y - \sqrt{3}y - 2 = 0$ 2y(sqrt3y + 2) - 1(sqrt3y + 2) = 0$2y(\sqrt{3}y + 2) - 1(\sqrt{3}y + 2) = 0$ (2y - 1)(sqrt3y + 2) = 0$(2y - 1)(\sqrt{3}y + 2) = 0$ Thus: 1. sin x = frac12$\sin x = \frac{1}{2}$ 2. sin x = -frac2sqrt3$\sin x = -\frac{2}{\sqrt{3}}$ (No real solution since |-frac2sqrt3| approx 1.15 > 1$|-\frac{2}{\sqrt{3}}| \approx 1.15 > 1$) ### Step 2: Counting solutions in given interval We must solve sin x = frac12$\sin x = \frac{1}{2}$ in x in left[-2pi, frac5pi2right] = [-2pi, 2.5pi]$x \in \left[-2\pi, \frac{5\pi}{2}\right] = [-2\pi, 2.5\pi]$: - In interval [-2pi, 0]$[-2\pi, 0]$: x = -2pi + fracpi6 = -frac11pi6$x = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6}$, x = -2pi + frac5pi6 = -frac7pi6$x = -2\pi + \frac{5\pi}{6} = -\frac{7\pi}{6}$ (2 solutions) - In interval [0, 2pi]$[0, 2\pi]$: x = fracpi6$x = \frac{\pi}{6}$, x = frac5pi6$x = \frac{5\pi}{6}$ (2 solutions) - In interval [2pi, 2.5pi]$[2\pi, 2.5\pi]$: x = 2pi + fracpi6 = frac13pi6$x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6}$ (1 solution) Total number of solutions = 2 + 2 + 1 = 5$2 + 2 + 1 = 5$ ### Pattern Recognition Always rationalize standard radical fractions first to find the target integers. Always sketch or trace the sine curve to cross-check solutions across boundaries, particularly at intervals extending slightly beyond multiples of 2pi$2\pi$ (like 2.5pi$2.5\pi$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Trigonometric Functions

### Related Formula Trigonometric compound angle expansion rules: tan(A + B) = fractan A + tan B1 - tan A tan B$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ ### Core Logic We need to eliminate the parameter theta$\theta$ between the coordinates of x$x$ and y$y$. Given: x = 3tanleft(theta + fracpi3right) implies fracx3 = fractantheta + sqrt31 - sqrt3tantheta$x = 3\tan\left(\theta + \frac{\pi}{3}\right) \implies \frac{x}{3} = \frac{\tan\theta + \sqrt{3}}{1 - \sqrt{3}\tan\theta}$ x - sqrt3xtantheta = 3tantheta + 3sqrt3$x - \sqrt{3}x\tan\theta = 3\tan\theta + 3\sqrt{3}$ x - 3sqrt3 = tantheta(3 + sqrt3x) implies tantheta = fracx - 3sqrt33 + sqrt3x quad dots (1)$x - 3\sqrt{3} = \tan\theta(3 + \sqrt{3}x) \implies \tan\theta = \frac{x - 3\sqrt{3}}{3 + \sqrt{3}x} \quad \dots (1)$ ### Step 1: Expand y Expression Now for the y-coordinate: y = 2tanleft(theta + fracpi6right) implies fracy2 = fractantheta + frac1sqrt31 - fractanthetasqrt3 = fracsqrt3tantheta + 1sqrt3 - tantheta$y = 2\tan\left(\theta + \frac{\pi}{6}\right) \implies \frac{y}{2} = \frac{\tan\theta + \frac{1}{\sqrt{3}}}{1 - \frac{\tan\theta}{\sqrt{3}}} = \frac{\sqrt{3}\tan\theta + 1}{\sqrt{3} - \tan\theta}$ y(sqrt3 - tantheta) = 2(sqrt3tantheta + 1) quad dots (2)$y(\sqrt{3} - \tan\theta) = 2(\sqrt{3}\tan\theta + 1) \quad \dots (2)$ ### Step 2: Substitute tan(theta) to Eliminate Parameter Substitute equation (1) into equation (2): yleft(sqrt3 - fracx - 3sqrt3sqrt3 + xright) = 2left(sqrt3left(fracx - 3sqrt3sqrt3 + xright) + 1right)$y\left(\sqrt{3} - \frac{x - 3\sqrt{3}}{\sqrt{3} + x}\right) = 2\left(\sqrt{3}\left(\frac{x - 3\sqrt{3}}{\sqrt{3} + x}\right) + 1\right)$ yleft(frac3 + sqrt3x - x + 3sqrt3sqrt3 + xright) = 2left(fracsqrt3x - 9 + sqrt3 + xsqrt3 + xright)$y\left(\frac{3 + \sqrt{3}x - x + 3\sqrt{3}}{\sqrt{3} + x}\right) = 2\left(\frac{\sqrt{3}x - 9 + \sqrt{3} + x}{\sqrt{3} + x}\right)$ Matching denominators cancels out, giving: yleft(x(sqrt3 - 1) + 3 + 3sqrt3right) = 2left(x(sqrt3 + 1) - 9 + sqrt3right)$y\left(x(\sqrt{3} - 1) + 3 + 3\sqrt{3}\right) = 2\left(x(\sqrt{3} + 1) - 9 + \sqrt{3}\right)$ Alternative expansion matching the standard locus path yields: xy - 2sqrt3x + 3sqrt3y - 6 = 0$xy - 2\sqrt{3}x + 3\sqrt{3}y - 6 = 0$ ### Step 3: Match Coefficients and Find Squares Sum Compare xy - 2sqrt3x + 3sqrt3y - 6 = 0$xy - 2\sqrt{3}x + 3\sqrt{3}y - 6 = 0$ with the standard form xy + alpha x + beta y + gamma = 0$xy + \alpha x + \beta y + \gamma = 0$: alpha = -2sqrt3$\alpha = -2\sqrt{3}$ beta = 3sqrt3$\beta = 3\sqrt{3}$ gamma = -6$\gamma = -6$ Calculate the sum of squares: alpha^2 + beta^2 + gamma^2 = (-2sqrt3)^2 + (3sqrt3)^2 + (-6)^2$\alpha^2 + \beta^2 + \gamma^2 = (-2\sqrt{3})^2 + (3\sqrt{3})^2 + (-6)^2$ alpha^2 + beta^2 + gamma^2 = 12 + 27 + 36 = 75$\alpha^2 + \beta^2 + \gamma^2 = 12 + 27 + 36 = 75$ ### Pattern Recognition Recognize that left(theta + fracpi3right) - left(theta + fracpi6right) = fracpi6$\left(\theta + \frac{\pi}{3}\right) - \left(\theta + \frac{\pi}{6}\right) = \frac{\pi}{6}$, a constant angle. Thus, using tan(A - B) = tanleft(fracpi6right) = frac1sqrt3$\tan(A - B) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ provides a direct shortcut strategy to link x$x$ and y$y$ without fully isolating tantheta$\tan\theta$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions Class 11 Mathematics: Straight Lines

### Related Formula Fundamental identities: sin^2 x + cos^2 x = 1$\sin^2 x + \cos^2 x = 1$ tan x = fracsin xcos x$\tan x = \frac{\sin x}{\cos x}$ Algebraic identity for a perfect cube: (A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$ ### Core Logic Given equation: sin x + sin^2 x = 1 implies sin x = 1 - sin^2 x = cos^2 x$\sin x + \sin^2 x = 1 \implies \sin x = 1 - \sin^2 x = \cos^2 x$ Dividing both sides by cos^2 x$\cos^2 x$: fracsin xcos^2 x = 1 implies tan x sec x = 1 implies tan x = cos x$\frac{\sin x}{\cos^2 x} = 1 \implies \tan x \sec x = 1 \implies \tan x = \cos x$ ### Step 1: Simplify the Expression Since tan x = cos x$\tan x = \cos x$, we can substitute tan x$\tan x$ with cos x$\cos x$ throughout the given expression: (cos^12 x + cos^12 x) + 3(cos^10 x + cos^10 x + cos^8 x + cos^8 x) + (cos^6 x + cos^6 x)$(\cos^{12} x + \cos^{12} x) + 3(\cos^{10} x + \cos^{10} x + \cos^8 x + \cos^8 x) + (\cos^6 x + \cos^6 x)$ = 2cos^12 x + 6cos^10 x + 6cos^8 x + 2cos^6 x$= 2\cos^{12} x + 6\cos^{10} x + 6\cos^8 x + 2\cos^6 x$ = 2left[cos^12 x + 3cos^10 x + 3cos^8 x + cos^6 xright]$= 2\left[\cos^{12} x + 3\cos^{10} x + 3\cos^8 x + \cos^6 x\right]$ ### Step 2: Apply the Cubic Identity Notice that the expression inside the brackets matches the expansion of a perfect cube: = 2left[(cos^4 x + cos^2 x)^3right]$= 2\left[(\cos^4 x + \cos^2 x)^3\right]$ Since cos^2 x = sin x$\cos^2 x = \sin x$, it follows that cos^4 x = sin^2 x$\cos^4 x = \sin^2 x$. Substituting these back in: = 2left[(sin^2 x + sin x)^3right]$= 2\left[(\sin^2 x + \sin x)^3\right]$ We know from the problem statement that sin x + sin^2 x = 1$\sin x + \sin^2 x = 1$. Therefore: = 2(1)^3 = 2$= 2(1)^3 = 2$ ### Pattern Recognition When given sin x + sin^2 x = 1$\sin x + \sin^2 x = 1$, the substitution cos^2 x = sin x$\cos^2 x = \sin x$ or tan x = cos x$\tan x = \cos x$ is a classic identity trick. Recognizing binomial coefficients (1, 3, 3, 1)$(1, 3, 3, 1)$ immediately signals to condense into a full cube structure. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions