$x\cos3\theta - 8y - 12z = 0$
$x\cos2\theta + 3y + 3z = 0$
$x + y + 3z = 0$
has a non-trivial solution, is equal to :
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.
The given system of equations is:
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 $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:
In the interval $\theta \in [0, 2\pi]$, the values are:
$ \theta = \frac{\pi}{2}, \frac{3\pi}{2} $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 $\cos\theta = c$:
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 $= 2\pi + 2\pi = 4\pi$.
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 :
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 ________.