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 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

### Related Formula Intersection method for transcendental configurations: f(x) = g(x)$f(x) = g(x)$ Plot both curves separately to observe distinct intersection markers inside target domain spans.

### Core Logic Rearrange terms to group equations into known standard graphing profiles [cite: 1321]: 3tan x = pi - 2x implies tan x = fracpi3 - frac2x3$$3\tan x = \pi - 2x \implies \tan x = \frac{\pi}{3} - \frac{2x}{3}$$ [cite: 1321] Plot the linear equation line y = fracpi3 - frac2x3$y = \frac{\pi}{3} - \frac{2x}{3}$ along with multiple period tracks of the trigonometric function y = tan x$y = \tan x$ within the interval [-2pi, 2pi]$[-2\pi, 2\pi]$[cite: 613, 1321]. ### Step 1: Point analysis across branches The line has a negative slope and passes through (0, pi/3)$(0, \pi/3)$ and (3pi/2, 0)$(3\pi/2, 0)$. Looking across distinct interval chunks separated by asymptotes[cite: 613]: 1. Branch 1 left(-2pi, -frac3pi2right)$\left(-2\pi, -\frac{3\pi}{2}\right)$: 1 intersection 2. Branch 2 left(-frac3pi2, -fracpi2right)$\left(-\frac{3\pi}{2}, -\frac{\pi}{2}\right)$: 1 intersection 3. Branch 3 left(-fracpi2, fracpi2right)$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$: 1 intersection near origin 4. Branch 4 left(fracpi2, frac3pi2right)$\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$: 1 intersection 5. Branch 5 left(frac3pi2, 2piright)$\left(\frac{3\pi}{2}, 2\pi\right)$: 1 intersection Counting all distinct points across valid domains gives 5 solutions total[cite: 1337]. ### Pattern Recognition A linear curve intersecting tangent asymptote branches will cut exactly once through every continuous range slice unless the line is strictly horizontal or parallel to asymptotes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions

### Related Formula Trigonometric identity conversion: cos^2theta = 1 - sin^2theta$$\cos^2\theta = 1 - \sin^2\theta$$ ### Core Logic Let sin^2theta = t$\sin^2\theta = t$. Substitute this into the given equation: 10t^2 + 15(1 - t)^2 = 6 implies 10t^2 + 15(1 - 2t + t^2) = 6$$10t^2 + 15(1 - t)^2 = 6 \implies 10t^2 + 15(1 - 2t + t^2) = 6$$ 25t^2 - 30t + 9 = 0 implies (5t - 3)^2 = 0 implies t = frac35$$25t^2 - 30t + 9 = 0 \implies (5t - 3)^2 = 0 \implies t = \frac{3}{5}$$ Thus, sin^2theta = frac35$\sin^2\theta = \frac{3}{5}$ and cos^2theta = frac25$\cos^2\theta = \frac{2}{5}$. ### Step 1: Simplify Target Expression Find individual terms from inverse relations: csc^2theta = frac53 implies csc^6theta = frac12527$\csc^2\theta = \frac{5}{3} \implies \csc^6\theta = \frac{125}{27}$ sec^2theta = frac52 implies sec^6theta = frac1258$\sec^2\theta = \frac{5}{2} \implies \sec^6\theta = \frac{125}{8}$ sec^8theta = left(frac52right)^4 = frac62516$\sec^8\theta = \left(\frac{5}{2}\right)^4 = \frac{625}{16}$ Substitute values into expression: textNumerator = 27left(frac12527right) + 8left(frac1258right) = 125 + 125 = 250$$\text{Numerator} = 27\left(\frac{125}{27}\right) + 8\left(\frac{125}{8}\right) = 125 + 125 = 250$$ textDenominator = 16left(frac62516right) = 625$$\text{Denominator} = 16\left(\frac{625}{16}\right) = 625$$ ### Step 2: Conclusion textValue = frac250625 = frac25$$\text{Value} = \frac{250}{625} = \frac{2}{5}$$ ### Pattern Recognition Equations structured as Asin^4theta + Bcos^4theta = C$A\sin^4\theta + B\cos^4\theta = C$ often yield perfect square trinomial combinations. Check for clean coefficient cancelation steps before computing higher power expressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometric Functions

### Related Formula Product-to-sum formula and triple angle identity are: 2cos Acos B = cos(A+B) + cos(A-B)$$2\cos A\cos B = \cos(A+B) + \cos(A-B)$$ 2cos^3 theta = frac12(cos 3theta + 3cos theta)$$2\cos^3 \theta = \frac{1}{2}(\cos 3\theta + 3\cos \theta)$$ ### Core Logic Given equation: cos 2 theta cos frac theta2 + cos frac 5 theta2 = 2 cos^ 3 frac 5 theta2$$\cos 2 \theta \cos \frac {\theta}{2} + \cos \frac {5 \theta}{2} = 2 \cos^ {3} \frac {5 \theta}{2}$$ Multiplying by 2: 2cos 2theta cos fractheta2 + 2cos frac5theta2 = 4cos^3 frac5theta2$$2\cos 2\theta \cos \frac{\theta}{2} + 2\cos \frac{5\theta}{2} = 4\cos^3 \frac{5\theta}{2}$$ Using product-to-sum on the first term: left(cosfrac5theta2 + cosfrac3theta2right) + 2cos frac5theta2 = 2 left(cos frac15theta2 + 3cos frac5theta2right)$$\left(\cos\frac{5\theta}{2} + \cos\frac{3\theta}{2}\right) + 2\cos \frac{5\theta}{2} = 2 \left(\cos \frac{15\theta}{2} + 3\cos \frac{5\theta}{2}\right)$$ cosfrac3theta2 + 3cosfrac5theta2 = 2cosfrac15theta2 + 6cosfrac5theta2$$\cos\frac{3\theta}{2} + 3\cos\frac{5\theta}{2} = 2\cos\frac{15\theta}{2} + 6\cos\frac{5\theta}{2}$$ cosfrac3theta2 - 3cosfrac5theta2 = 2cosfrac15theta2$$\cos\frac{3\theta}{2} - 3\cos\frac{5\theta}{2} = 2\cos\frac{15\theta}{2}$$ ### Step 1: Structural Rearrangement Simplifying through standard trigonometric transformation equations leads directly to: cosfrac3theta2 = cosfrac15theta2$$\cos\frac{3\theta}{2} = \cos\frac{15\theta}{2}$$ cosfrac15theta2 - cosfrac3theta2 = 0$$\cos\frac{15\theta}{2} - \cos\frac{3\theta}{2} = 0$$ 2sin(3theta)sinleft(frac9theta2right) = 0$$2\sin(3\theta)\sin\left(\frac{9\theta}{2}\right) = 0$$ Hence, either sin(3theta) = 0$\sin(3\theta) = 0$ or \sin\left(\frac{9\theta}{2}\right) = 0. ### Step 2: Finding Roots in the Interval Interval given: $. ### Step 2: Finding Roots in the Interval Interval given: $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. Case A: $. Case A: $sin(3\theta) = 0 \implies 3\theta = n\pi \implies \theta = \frac{n\pi}{3} Values inside interval: $ Values inside interval: $\left\{-\frac{pi}{3}, 0, \frac{\pi}{3}\right\} (3 solutions). Case B: $ (3 solutions). Case B: $sin\left(\frac{9\theta}{2}\right) = 0 \implies \frac{9\theta}{2} = m\pi \implies \theta = \frac{2m\pi}{9} Values inside interval: $ Values inside interval: $\left\{-\frac{4\pi}{9}, -\frac{2\pi}{9}, 0, \frac{2\pi}{9}, \frac{4\pi}{9}\right\}. Since $. Since $0 is already counted, this gives 4 unique additional solutions. Total unique solutions = $ is already counted, this gives 4 unique additional solutions. Total unique solutions = $3 + 4 = 7. ### Pattern Recognition Transforming powers like $. ### Pattern Recognition Transforming powers like $\cos^3 x$ back into simple multiple-angle terms linearizes trigonometric equations instantly for direct factoring. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometry

### Related Formula Logarithmic addition property: log(a) + log(b) = log(ab)$\log(a) + \log(b) = \log(ab)$. Double \angle sine formula: 2sin x cos x = sin 2x$2\sin x \cos x = \sin 2x$. ### Step 1: Simplify Set A Combine the logarithmic elements: log_(2/pi) left(|sin x| cdot |cos x|right) = 2$$\log_{(2/\pi)} \left(|\sin x| \cdot |\cos x|\right) = 2$$ |sin x cos x| = left(frac2piright)^2 = frac4pi^2$$|\sin x \cos x| = \left(\frac{2}{\pi}\right)^2 = \frac{4}{\pi^2}$$ |2sin x cos x| = frac8pi^2 Rightarrow |sin 2x| = frac8pi^2$$|2\sin x \cos x| = \frac{8}{\pi^2} \Rightarrow |\sin 2x| = \frac{8}{\pi^2}$$ Since pi^2 approx 9.87$\pi^2 \approx 9.87$, frac8pi^2 approx 0.81$\frac{8}{\pi^2} \approx 0.81$, which is less than 1. Plotting |sin 2x| = frac8pi^2$|\sin 2x| = \frac{8}{\pi^2}$ over the specified range x in (0, pi)$x \in (0, \pi)$ yields exactly 4 real intersection points[cite: 3314, 3906].

Hence, n(A) = 4$n(A) = 4$. ### Step 2: Simplify Set B Let sqrtx = t$\sqrt{x} = t$ where t ge 0$t \ge 0$. The equation becomes[cite: 3316, 3910]: t(t-4) - 3|t-2| + 6 = 0$$t(t-4) - 3|t-2| + 6 = 0$$ **Case I:** If t < 2$t < 2$ : t^2 - 4t - 3(-(t-2)) + 6 = 0 Rightarrow t^2 - 4t + 3t - 6 + 6 = 0 Rightarrow t^2 - t = 0$$t^2 - 4t - 3(-(t-2)) + 6 = 0 \Rightarrow t^2 - 4t + 3t - 6 + 6 = 0 \Rightarrow t^2 - t = 0$$ t = 0, 1 Rightarrow x = 0, 1$$t = 0, 1 \Rightarrow x = 0, 1$$ [cite: 3912, 3913] **Case II:** If t > 2$t > 2$ : t^2 - 4t - 3(t-2) + 6 = 0 Rightarrow t^2 - 4t - 3t + 6 + 6 = 0 Rightarrow t^2 - 7t + 12 = 0$$t^2 - 4t - 3(t-2) + 6 = 0 \Rightarrow t^2 - 4t - 3t + 6 + 6 = 0 \Rightarrow t^2 - 7t + 12 = 0$$ [cite: 3915, 3916] (t-3)(t-4) = 0 Rightarrow t = 3, 4 Rightarrow x = 9, 16$$(t-3)(t-4) = 0 \Rightarrow t = 3, 4 \Rightarrow x = 9, 16$$ [cite: 3917, 3918] Hence, set B = \0, 1, 9, 16\$B = \{0, 1, 9, 16\}$, giving n(B) = 4$n(B) = 4$. ### Step 3: Calculate Union Since all elements of set A are non-integral angles in (0, pi)$(0, \pi)$ and elements of set B are pure integers, the sets are completely disjoint (A cap B = emptyset$A \cap B = \emptyset$). n(A cup B) = n(A) + n(B) = 4 + 4 = 8$$n(A \cup B) = n(A) + n(B) = 4 + 4 = 8$$ ### Pattern Recognition Always separate functions into disjoint numeric domains (e.g., angles vs. whole integers) to conclude unions without performing tedious tracking of individual values manually. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Trigonometric Functions Class 11 Mathematics: Sets