All Exams Test series for 1 year @ ₹349 only
Question

The sum of all possible values of $\theta \in [0, 2\pi]$, for which the system of equations :
$x\cos3\theta - 8y - 12z = 0$
$x\cos2\theta + 3y + 3z = 0$
$x + y + 3z = 0$
has a non-trivial solution, is equal to :

The correct answer is
$4\pi$

System Equations Non-Trivial Solution Condition

A system of homogeneous linear equations has a non-trivial solution if and only if the determinant of the coefficient matrix is equal to zero.

Coefficient Matrix and Determinant Calculation

The given system of equations is:

  • $x\cos3\theta - 8y - 12z = 0$
  • $x\cos2\theta + 3y + 3z = 0$
  • $x + y + 3z = 0$

The coefficient matrix $A$ is:

$ A = \begin{pmatrix} \cos3\theta & -8 & -12 \\ \cos2\theta & 3 & 3 \\ 1 & 1 & 3 \end{pmatrix} $

For a non-trivial solution, $\det(A) = 0$. Calculating the determinant:

$ \det(A) = \cos3\theta(3 \cdot 3 - 3 \cdot 1) - (-8)(\cos2\theta \cdot 3 - 3 \cdot 1) + (-12)(\cos2\theta \cdot 1 - 3 \cdot 1) $ $ = \cos3\theta(9 - 3) + 8(3\cos2\theta - 3) - 12(\cos2\theta - 3) $ $ = 6\cos3\theta + 24\cos2\theta - 24 - 12\cos2\theta + 36 $ $ = 6\cos3\theta + 12\cos2\theta + 12 $

Setting the determinant to zero:

$ 6\cos3\theta + 12\cos2\theta + 12 = 0 $

Divide by 6:

$ \cos3\theta + 2\cos2\theta + 2 = 0 $

Solving Trigonometric Equation for Theta

Use the identities $\cos3\theta = 4\cos^3\theta - 3\cos\theta$ and $\cos2\theta = 2\cos^2\theta - 1$. Let $c = \cos\theta$. Substitute these into the equation:

$ (4c^3 - 3c) + 2(2c^2 - 1) + 2 = 0 $ $ 4c^3 - 3c + 4c^2 - 2 + 2 = 0 $ $ 4c^3 + 4c^2 - 3c = 0 $

Factor out $c$:

$ c(4c^2 + 4c - 3) = 0 $

This gives two possibilities:

  1. $c = \cos\theta = 0$
  2. $4c^2 + 4c - 3 = 0$

Case 1: $\cos\theta = 0$

In the interval $\theta \in [0, 2\pi]$, the values are:

$ \theta = \frac{\pi}{2}, \frac{3\pi}{2} $

Case 2: $4c^2 + 4c - 3 = 0$

Solve the quadratic equation for $c$ using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$ c = \frac{-4 \pm \sqrt{4^2 - 4(4)(-3)}}{2(4)} = \frac{-4 \pm \sqrt{16 + 48}}{8} = \frac{-4 \pm \sqrt{64}}{8} = \frac{-4 \pm 8}{8} $

The possible values for $c$ are:

$ c_1 = \frac{-4 + 8}{8} = \frac{4}{8} = \frac{1}{2} $ $ c_2 = \frac{-4 - 8}{8} = \frac{-12}{8} = -\frac{3}{2} $

Now, consider $\cos\theta = c$:

  • If $\cos\theta = \frac{1}{2}$: In the interval $\theta \in [0, 2\pi]$, the values are $\theta = \frac{\pi}{3}, \frac{5\pi}{3}$.
  • If $\cos\theta = -\frac{3}{2}$: This is impossible since $-1 \le \cos\theta \le 1$.

Sum of Possible Theta Values

The possible values for $\theta$ in the interval $[0, 2\pi]$ are $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{3}, \frac{5\pi}{3}$.

Sum $= \frac{\pi}{2} + \frac{3\pi}{2} + \frac{\pi}{3} + \frac{5\pi}{3}$

Sum $= \left(\frac{\pi}{2} + \frac{3\pi}{2}\right) + \left(\frac{\pi}{3} + \frac{5\pi}{3}\right)$

Sum $= \frac{4\pi}{2} + \frac{6\pi}{3}$

Sum $= 2\pi + 2\pi = 4\pi$.

Was this answer helpful?

Similar Questions

  1. Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
  2. Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
  3. Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
  4. If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
  5. Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
  6. The number of elements in the relation $R = \{(x, y) : 4x^2 + y^2 < 52, x, y \in \mathbb{Z}\}$ is
  7. Let $S = \{z \in \mathbb{C} : 4z^2 + \bar{z} = 0\}$. Then $\sum_{z \in S} |z|^2$ is equal to :
  8. Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :

  9. Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.

  10. Let $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ up to 13 terms. If $13S = \frac{2^k}{n!}, k \in \mathbb{N}$, then $n + k$ is equal to

Important Questions from Algebra

  1. Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
  2. Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
  3. Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
  4. If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
  5. Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
Need Expert Advice?
More Questions from JEE Main

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App