Solution & Explanation
### 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
More Trigonometric Functions Previous-Year Questions
Q59
jee_main_2025_02_april_evening
Trigonometric Equations
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
Solution
### 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
Q
jee_main_2025_02_april_morning
Trigonometric Equations
If theta in [-2pi, 2pi]$\theta \in [-2\pi, 2\pi]$, then the number of solutions of 2sqrt2cos^2theta + (2 - sqrt6)costheta - sqrt3 = 0$2\sqrt{2}\cos^2\theta + (2 - \sqrt{6})\cos\theta - \sqrt{3} = 0$, is equal to:
- A. 12$12$
- B. 6$6$
- C. 8$8$
- D. 10$10$
Solution
### 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
Q66
jee_main_2025_03_april_evening
Trigonometric Equations
The number of solutions of equation (4 - sqrt3)sin x - 2sqrt3cos^2 x = -frac41 + sqrt3$(4 - \sqrt{3})\sin x - 2\sqrt{3}\cos^2 x = -\frac{4}{1 + \sqrt{3}}$, x in left[-2pi, frac5pi2right]$x \in \left[-2\pi, \frac{5\pi}{2}\right]$ is
- A. 4$4$
- B. 3$3$
- C. 6$6$
- D. 5$5$
Solution
### 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
Q58
jee_main_2025_07_april_morning
Trigonometric Equations and Locus
If for theta in left[-fracpi3, 0right]$\theta \in \left[-\frac{\pi}{3}, 0\right]$ , the points left(mathrmx,mathrmyright) = left(3tan left(theta +fracpi3right),2tan left(theta +fracpi6right)right)$\left(\mathrm{x},\mathrm{y}\right) = \left(3\tan \left(\theta +\frac{\pi}{3}\right),2\tan \left(\theta +\frac{\pi}{6}\right)\right)$ lie on mathrmxy + alpha mathrmx + beta mathrmy + gamma = 0,$\mathrm{xy} + \alpha \mathrm{x} + \beta \mathrm{y} + \gamma = 0,$ then alpha^2 +beta^2 +gamma^2$\alpha^2 +\beta^2 +\gamma^2$ is equal to:
- A. 80$80$
- B. 72$72$
- C. 96$96$
- D. 75$75$
Solution
### 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
Q52
jee_main_2025_29_jan_evening
Trigonometric Identities
If sin x + sin^2 x = 1$\sin x + \sin^2 x = 1$, x in left(0, fracpi2right)$x \in \left(0, \frac{\pi}{2}\right)$, then
(cos^12 x + tan^12 x) + 3 (cos^10 x + tan^10 x + cos^8 x + tan^8 x) + (cos^6 x + tan^6 x)$(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^8 x + \tan^8 x) + (\cos^6 x + \tan^6 x)$ is equal to
- A. 4$4$
- B. 3$3$
- C. 2$2$
- D. 1$1$
Solution
### 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
