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 — Page 3
Q4
jee_main_2024_01_february_morning
Trigonometric Identities
If
tan A=frac1sqrtx(x^2+x+1)$\tan A=\frac{1}{\sqrt{x(x^{2}+x+1)}}$,
tan B=fracsqrtxsqrtx^2+x+1$\tan B=\frac{\sqrt{x}}{\sqrt{x^{2}+x+1}}$ and
tan C=(x^-3+x^-2+x^-1)^frac12, 0$\tan C=(x^{-3}+x^{-2}+x^{-1})^{\frac{1}{2}}, 0, then A+B$A+B$ is equal to:
- A. C$C$
- B. pi-C$\pi-C$
- C. 2pi-C$2\pi-C$
- D. fracpi2-C$\frac{\pi}{2}-C$
Solution
### Related Formula
Trigonometric Addition Identity for tangent:
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
Given expressions for tan A$\tan A$ and tan B$\tan B$:
tan(A+B) = fracfrac1sqrtx(x^2+x+1) + fracsqrtxsqrtx^2+x+11 - left(frac1sqrtx(x^2+x+1)right) cdot left(fracsqrtxsqrtx^2+x+1right) $\tan(A+B) = \frac{\frac{1}{\sqrt{x(x^2+x+1)}} + \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1 - \left(\frac{1}{\sqrt{x(x^2+x+1)}}\right) \cdot \left(\frac{\sqrt{x}}{\sqrt{x^2+x+1}}\right)} $
### Step 1: Simplify the Compound Tangent Formula
Simplify the numerator:
textNumerator = frac1 + xsqrtxsqrtx^2+x+1$\text{Numerator} = \frac{1 + x}{\sqrt{x}\sqrt{x^2+x+1}}$
Simplify the denominator:
textDenominator = 1 - frac1x^2+x+1 = fracx^2+x+1-1x^2+x+1 = fracx^2+xx^2+x+1 = fracx(x+1)x^2+x+1$\text{Denominator} = 1 - \frac{1}{x^2+x+1} = \frac{x^2+x+1-1}{x^2+x+1} = \frac{x^2+x}{x^2+x+1} = \frac{x(x+1)}{x^2+x+1}$
Now put them together:
tan(A+B) = fracfrac1+xsqrtxsqrtx^2+x+1fracx(x+1)x^2+x+1 = frac(1+x)(x^2+x+1)sqrtxsqrtx^2+x+1 cdot x(x+1) $\tan(A+B) = \frac{\frac{1+x}{\sqrt{x}\sqrt{x^2+x+1}}}{\frac{x(x+1)}{x^2+x+1}} = \frac{(1+x)(x^2+x+1)}{\sqrt{x}\sqrt{x^2+x+1} \cdot x(x+1)} $
### Step 2: Compare with tan C
Cancelling out (1+x)$(1+x)$ and matching root expressions:
tan(A+B) = fracsqrtx^2+x+1xsqrtx $\tan(A+B) = \frac{\sqrt{x^2+x+1}}{x\sqrt{x}} $
Now evaluate tan C$\tan C$:
tan C = sqrtfrac1x^3 + frac1x^2 + frac1x = sqrtfrac1+x+x^2x^3 = fracsqrtx^2+x+1xsqrtx $\tan C = \sqrt{\frac{1}{x^3} + \frac{1}{x^2} + \frac{1}{x}} = \sqrt{\frac{1+x+x^2}{x^3}} = \frac{\sqrt{x^2+x+1}}{x\sqrt{x}} $
Since tan(A+B) = tan C$\tan(A+B) = \tan C$ and both arguments are in acute range:
A+B = C$A+B = C$
### Pattern Recognition
Sees: Multi-variable algebraic rational terms involving square roots.
Shortcut: If algebraic tracking feels complicated, substitute a simple valid number like x=1$x=1$ to evaluate coefficients dynamically: tan A = frac1sqrt3$\tan A = \frac{1}{\sqrt{3}}$, tan B = frac1sqrt3 implies A=30^circ, B=30^circ implies A+B=60^circ$\tan B = \frac{1}{\sqrt{3}} \implies A=30^{\circ}, B=30^{\circ} \implies A+B=60^{\circ}$. Then tan C = sqrt1+1+1 = sqrt3 implies C=60^circ$\tan C = \sqrt{1+1+1} = \sqrt{3} \implies C=60^{\circ}$. Thus A+B=C$A+B=C$ holds instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Trigonometric Functions
Class 10 Mathematics: Algebraic Identities
Q7
jee_main_2024_29_january_evening
Trigonometric Equations
The sum of the solutions x in mathbbR$x \in \mathbb{R}$ of the equation frac3 cos 2 x + cos^3 2 xcos^6 x - sin^6 x = x^3 - x^2 + 6$\frac{3 \cos 2 x + \cos^3 2 x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6$ is
Solution
### Related Formula
cos^6 x - sin^6 x = (cos^2 x - sin^2 x)(cos^4 x + cos^2 x sin^2 x + sin^4 x) = cos 2x (1 - sin^2 x cos^2 x)$\cos^6 x - \sin^6 x = (\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) = \cos 2x (1 - \sin^2 x \cos^2 x)$
### Core Logic
Let us simplify the LHS expression:
textLHS = fraccos 2x (3 + cos^2 2x)cos 2x (1 - sin^2 x cos^2 x)$\text{LHS} = \frac{\cos 2x (3 + \cos^2 2x)}{\cos 2x (1 - \sin^2 x \cos^2 x)}$
Assuming cos 2x neq 0$\cos 2x \neq 0$:
textLHS = frac3 + cos^2 2x1 - frac14sin^2 2x = frac4(3 + cos^2 2x)4 - sin^2 2x$\text{LHS} = \frac{3 + \cos^2 2x}{1 - \frac{1}{4}\sin^2 2x} = \frac{4(3 + \cos^2 2x)}{4 - \sin^2 2x}$
Since sin^2 2x = 1 - cos^2 2x$\sin^2 2x = 1 - \cos^2 2x$, the denominator becomes:
4 - (1 - cos^2 2x) = 3 + cos^2 2x$4 - (1 - \cos^2 2x) = 3 + \cos^2 2x$
Therefore:
textLHS = frac4(3 + cos^2 2x)3 + cos^2 2x = 4$\text{LHS} = \frac{4(3 + \cos^2 2x)}{3 + \cos^2 2x} = 4$
### Step 1: Solving the Algebraic Equation
Equating LHS to RHS:
4 = x^3 - x^2 + 6 implies x^3 - x^2 + 2 = 0$4 = x^3 - x^2 + 6 \implies x^3 - x^2 + 2 = 0$
By inspection, x = -1$x = -1$ is a root:
(-1)^3 - (-1)^2 + 2 = -1 - 1 + 2 = 0$(-1)^3 - (-1)^2 + 2 = -1 - 1 + 2 = 0$
Factoring out (x + 1)$(x + 1)$:
(x + 1)(x^2 - 2x + 2) = 0$(x + 1)(x^2 - 2x + 2) = 0$
For the quadratic factor x^2 - 2x + 2 = 0$x^2 - 2x + 2 = 0$, the discriminant is D = (-2)^2 - 4(1)(2) = -4 < 0$D = (-2)^2 - 4(1)(2) = -4 < 0$, yielding no real roots.
Thus, the only real solution is x = -1$x = -1$, and its sum is -1$-1$.
### Pattern Recognition
Complicated mixed expressions of trigonometric fractions often collapse into simple constants upon identity transformations. Look for factorization templates like a^3 - b^3$a^3 - b^3$ or a^6 - b^6$a^6 - b^6$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Trigonometric Functions
Class 12 Mathematics: Polynomial Equations
Q27
jee_main_2024_27_jan_morning
Multiple Angles and Equations
Let the set of all ain R$a\in R$ such that the equation cos 2x+asin x=2a-7$\cos 2x+a\sin x=2a-7$ has a solution be [p, q]$[p, q]$ and r=tan 9^circ-tan 27^circ-frac1cot 63^circ+tan 81^circ$r=\tan 9^{\circ}-\tan 27^{\circ}-\frac{1}{\cot 63^{\circ}}+\tan 81^{\circ}$, then pqr$pqr$ is equal to:
Numerical Answer. Answer: 48 to 48
Solution
### Related Formula
cos 2x = 1 - 2sin^2 x$\cos 2x = 1 - 2\sin^2 x$
tan theta + cot theta = frac2sin 2theta$\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}$
### Core Logic
Transform the trigonometric equation into a quadratic in terms of sin x$\sin x$:
(1 - 2sin^2 x) + asin x = 2a - 7$(1 - 2\sin^2 x) + a\sin x = 2a - 7$
2sin^2 x - asin x + 2a - 8 = 0$2\sin^2 x - a\sin x + 2a - 8 = 0$
Factorizing the quadratic:
2sin^2 x - 4sin x - (a-4)sin x + 2(a-4) = 0$2\sin^2 x - 4\sin x - (a-4)\sin x + 2(a-4) = 0$
2sin x(sin x - 2) - (a-4)(sin x - 2) = 0$2\sin x(\sin x - 2) - (a-4)(\sin x - 2) = 0$
(sin x - 2)(2sin x - (a-4)) = 0$(\sin x - 2)(2\sin x - (a-4)) = 0$
### Step 1: Finding bounds for a
Since sin x = 2$\sin x = 2$ has no real solution, we must have:
sin x = fraca-42$\sin x = \frac{a-4}{2}$
For this to have a solution, the root must lie in the standard domain of sine:
-1 le fraca-42 le 1$-1 \le \frac{a-4}{2} \le 1$
-2 le a-4 le 2$-2 \le a-4 \le 2$
2 le a le 6$2 \le a \le 6$
Thus, the solution set is [p, q] = [2, 6]$[p, q] = [2, 6]$, meaning p = 2$p = 2$ and q = 6$q = 6$.
### Step 2: Evaluating r
Evaluate r = tan 9^circ - tan 27^circ - frac1cot 63^circ + tan 81^circ$r = \tan 9^{\circ} - \tan 27^{\circ} - \frac{1}{\cot 63^{\circ}} + \tan 81^{\circ}$.
Using complementary angles (tan(90 - theta) = cot theta$\tan(90 - \theta) = \cot \theta$):
tan 81^circ = cot 9^circ$\tan 81^{\circ} = \cot 9^{\circ}$
frac1cot 63^circ = tan 63^circ = cot 27^circ$\frac{1}{\cot 63^{\circ}} = \tan 63^{\circ} = \cot 27^{\circ}$
Substitute these in:
r = (tan 9^circ + cot 9^circ) - (tan 27^circ + cot 27^circ)$r = (\tan 9^{\circ} + \cot 9^{\circ}) - (\tan 27^{\circ} + \cot 27^{\circ})$
Apply the formula tan theta + cot theta = frac2sin 2theta$\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}$:
r = frac2sin 18^circ - frac2sin 54^circ$r = \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$
We know sin 18^circ = fracsqrt5-14$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and sin 54^circ = cos 36^circ = fracsqrt5+14$\sin 54^{\circ} = \cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.
r = frac8sqrt5-1 - frac8sqrt5+1 = 8 left[ fracsqrt5+1 - (sqrt5-1)(sqrt5-1)(sqrt5+1) right]$r = \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1} = 8 \left[ \frac{\sqrt{5}+1 - (\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} \right]$
r = 8 left[ frac24 right] = 4$r = 8 \left[ \frac{2}{4} \right] = 4$
### Step 3: Final Output Calculation
We need the value of pqr$pqr$:
pqr = 2 times 6 times 4 = 48$pqr = 2 \times 6 \times 4 = 48$
### Pattern Recognition
Converting mixed trig degrees like 9, 27, 63, 81 entirely into cot/tan pairs ALWAYS drops them into the frac2sin 2theta$\frac{2}{\sin 2\theta}$ double-angle trap, bringing them natively to 18 and 54 degrees.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Trigonometric Functions
Q11
jee_main_2024_29_jan_morning
Trigonometric Equations
If alpha$\alpha$, -fracpi2 lt alpha lt fracpi2$-\frac{\pi}{2} \lt \alpha \lt \frac{\pi}{2}$ is the solution of 4costheta+5sintheta=1$4\cos\theta+5\sin\theta=1$, then the value of tanalpha$\tan\alpha$ is
- A. frac10-sqrt106$\frac{10-\sqrt{10}}{6}$
- B. frac10-sqrt1012$\frac{10-\sqrt{10}}{12}$
- C. fracsqrt10-1012$\frac{\sqrt{10}-10}{12}$
- D. fracsqrt10-106$\frac{\sqrt{10}-10}{6}$
Solution
### Related Formula
sec^2theta - tan^2theta = 1$\sec^2\theta - \tan^2\theta = 1$
If acostheta + bsintheta = c$a\cos\theta + b\sin\theta = c$, dividing by costheta$\cos\theta$ transforms the equation into a quadratic in terms of tantheta$\tan\theta$ and sectheta$\sec\theta$.
### Core Logic
Given the equation:
4costheta + 5sintheta = 1$4\cos\theta + 5\sin\theta = 1$
Divide the entire equation by costheta$\cos\theta$:
4 + 5tantheta = sectheta$4 + 5\tan\theta = \sec\theta$
Square both sides to convert the sectheta$\sec\theta$ into a tantheta$\tan\theta$ expression:
(4 + 5tantheta)^2 = sec^2theta$(4 + 5\tan\theta)^2 = \sec^2\theta$
16 + 25tan^2theta + 40tantheta = 1 + tan^2theta$16 + 25\tan^2\theta + 40\tan\theta = 1 + \tan^2\theta$
Rearranging into a standard quadratic equation in terms of tantheta$\tan\theta$:
24tan^2theta + 40tantheta + 15 = 0$24\tan^2\theta + 40\tan\theta + 15 = 0$
### Step 1: Apply Quadratic Formula
Solve for tantheta$\tan\theta$ using the quadratic formula:
tantheta = frac-40 pm sqrt1600 - 4(24)(15)2(24)$\tan\theta = \frac{-40 \pm \sqrt{1600 - 4(24)(15)}}{2(24)}$
tantheta = frac-40 pm sqrt1600 - 144048$\tan\theta = \frac{-40 \pm \sqrt{1600 - 1440}}{48}$
tantheta = frac-40 pm sqrt16048$\tan\theta = \frac{-40 \pm \sqrt{160}}{48}$
tantheta = frac-40 pm 4sqrt1048$\tan\theta = \frac{-40 \pm 4\sqrt{10}}{48}$
tantheta = frac-10 pm sqrt1012$\tan\theta = \frac{-10 \pm \sqrt{10}}{12}$
This gives two possible values:
tantheta = frac-10 + sqrt1012 quad textand quad tantheta = -left(frac10 + sqrt1012right)$\tan\theta = \frac{-10 + \sqrt{10}}{12} \quad \text{and} \quad \tan\theta = -\left(\frac{10 + \sqrt{10}}{12}\right)$
### Step 2: Check Extraneous Roots
When we squared the equation 4 + 5tantheta = sectheta$4 + 5\tan\theta = \sec\theta$, we introduced the possibility of extraneous roots where sectheta$\sec\theta$ might be strictly negative while 4 + 5tantheta$4 + 5\tan\theta$ is negative, but alpha in (-pi/2, pi/2)$\alpha \in (-\pi/2, \pi/2)$ restricts cosalpha gt 0$\cos\alpha \gt 0$, hence secalpha gt 0$\sec\alpha \gt 0$.
For secalpha$\sec\alpha$ to be positive, 4 + 5tanalpha gt 0$4 + 5\tan\alpha \gt 0$.
If tanalpha = -frac10 + sqrt1012$\tan\alpha = -\frac{10 + \sqrt{10}}{12}$ (approx -1.09$-1.09$):
4 + 5(-1.09) = 4 - 5.45 = -1.45 lt 0$4 + 5(-1.09) = 4 - 5.45 = -1.45 \lt 0$
This contradicts secalpha gt 0$\sec\alpha \gt 0$. Hence, this root is rejected.
Therefore, the only valid solution is:
tanalpha = fracsqrt10 - 1012$\tan\alpha = \frac{\sqrt{10} - 10}{12}$
### Pattern Recognition
Whenever you square a trigonometric equation (like converting sec$\sec$ to tan$\tan$), always map the proposed roots back to the domain limits to prune out extraneous negative parity roots.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Trigonometric Functions
Q2
jee_main_2024_30_january_evening
Compound Angles
For alpha, beta in left(0, fracpi2right)$\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$ , let 3sin (alpha + beta) = 2sin (alpha - beta)$3\sin (\alpha + \beta) = 2\sin (\alpha - \beta)$ and a real number k$k$ be such that tan alpha = ktan beta$\tan \alpha = k\tan \beta$ . Then the value of k$k$ is equal to:
- A. -frac23$-\frac{2}{3}$
- B. -5$-5$
- C. frac23$\frac{2}{3}$
- D. 5$5$
Solution
### 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
