Question

If system of equations

$x + (sin\alpha)y + (sin^2\alpha)z = 0,$

$x + (cos\alpha)y + (cos^2\alpha)z = 0$

$x + (sin2\alpha)y + (sin^2 2\alpha)z = 0$

has non trivial solutions, then number of distinct values of $\alpha$ (where $\alpha \in [0,\pi]$), is

Answer 1

7

Answer 2

The given system of linear equations is:

$x + (\sin\alpha)y + (\sin^2\alpha)z = 0$

$x + (\cos\alpha)y + (\cos^2\alpha)z = 0$

$x + (\sin2\alpha)y + (\sin^2 2\alpha)z = 0$

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

The coefficient matrix is:

$A = \begin{pmatrix} 1 & \sin\alpha & \sin^2\alpha \\ 1 & \cos\alpha & \cos^2\alpha \\ 1 & \sin2\alpha & \sin^2 2\alpha \end{pmatrix}$

The determinant of this matrix is of the Vandermonde form $\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = (b-a)(c-a)(c-b)$, where $a = \sin\alpha$, $b = \cos\alpha$, and $c = \sin2\alpha$.

So, the determinant of the coefficient matrix is:

$\det(A) = (\cos\alpha - \sin\alpha)(\sin2\alpha - \sin\alpha)(\sin2\alpha - \cos\alpha)$

For the system to have non-trivial solutions, the determinant must be zero:

$(\cos\alpha - \sin\alpha)(\sin2\alpha - \sin\alpha)(\sin2\alpha - \cos\alpha) = 0$

This equation holds if any of the factors are zero. We need to find the distinct values of $\alpha$ in the interval $[0, \pi]$ that satisfy this condition.

Case 1: $\cos\alpha - \sin\alpha = 0$

$\cos\alpha = \sin\alpha$

$\tan\alpha = 1$

In the interval $[0, \pi]$, the solution is $\alpha = \frac{\pi}{4}$.

Case 2: $\sin2\alpha - \sin\alpha = 0$

$2\sin\alpha\cos\alpha - \sin\alpha = 0$

$\sin\alpha(2\cos\alpha - 1) = 0$

This gives two sub-cases:

a) $\sin\alpha = 0$

In the interval $[0, \pi]$, the solutions are $\alpha = 0$ and $\alpha = \pi$.

b) $2\cos\alpha - 1 = 0 \implies \cos\alpha = \frac{1}{2}$

In the interval $[0, \pi]$, the solution is $\alpha = \frac{\pi}{3}$.

Case 3: $\sin2\alpha - \cos\alpha = 0$

$2\sin\alpha\cos\alpha - \cos\alpha = 0$

$\cos\alpha(2\sin\alpha - 1) = 0$

This gives two sub-cases:

a) $\cos\alpha = 0$

In the interval $[0, \pi]$, the solution is $\alpha = \frac{\pi}{2}$.

b) $2\sin\alpha - 1 = 0 \implies \sin\alpha = \frac{1}{2}$

In the interval $[0, \pi]$, the solutions are $\alpha = \frac{\pi}{6}$ and $\alpha = \frac{5\pi}{6}$.

Combining all the distinct values obtained from the three cases within the interval $[0, \pi]$:

From Case 1: $\frac{\pi}{4}$

From Case 2: $0, \pi, \frac{\pi}{3}$

From Case 3: $\frac{\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}$

The set of all distinct values of $\alpha$ in $[0, \pi]$ is $\{0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{5\pi}{6}, \pi\}$.

These values are $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ, 150^\circ, 180^\circ$.

All these values are distinct and lie within the interval $[0, \pi]$.

The number of distinct values of $\alpha$ is 7.