Let, α and β be real. Find the set of all values of β for which the system of equation βx + sin α*y + cosα*z = 0, x + cosα * y + sinα * z = 0 , -x + sinα*y – cosα * z = 0 has a non-trivial solution. For β = 1 what are all values of α?

The correct option is: 2α = 2nπ ± π/4 + π/4

Step-by-Step Explanation:

We are given a system of homogeneous linear equations in terms of variables $x$, $y$, and $z$:

$$\begin{array}{l}\beta x+\sin \alpha \cdot y+\cos \alpha \cdot z=0\\ x+\cos \alpha \cdot y+\sin \alpha \cdot z=0\\ -x+\sin \alpha \cdot y-\cos \alpha \cdot z=0\end{array}$$

For a system of homogeneous linear equations to have a non-trivial solution (other than $x=y=z=0$), the determinant of the coefficient matrix must be equal to zero.

Let us write the determinant of the coefficients:

$$\left|\begin{array}{ccc}\beta & \sin \alpha & \cos \alpha \\ 1& \cos \alpha & \sin \alpha \\ -1& \sin \alpha & -\cos \alpha \end{array}\right|=0$$

Now, we expand the determinant along the first row:

$$\beta \left[\cos \alpha \left(-\cos \alpha \right)-{\sin }^2\alpha \right]-\sin \alpha \left[1\left(-\cos \alpha \right)-\sin \alpha \left(-1\right)\right]+\cos \alpha \left[1\left(\sin \alpha \right)-\cos \alpha \left(-1\right)\right]=0$$

Simplifying the terms inside each bracket:

$$\beta \left[-{\cos }^2\alpha -{\sin }^2\alpha \right]-\sin \alpha \left[-\cos \alpha +\sin \alpha \right]+\cos \alpha \left[\sin \alpha +\cos \alpha \right]=0$$

Using the trigonometric identity ${\sin }^2\alpha +{\cos }^2\alpha =1$:

$$-\beta -{\sin }^2\alpha +\sin \alpha \cos \alpha +\sin \alpha \cos \alpha +{\cos }^2\alpha =0$$

Rearranging the terms:

$$-\beta +\left({\cos }^2\alpha -{\sin }^2\alpha \right)+2\sin \alpha \cos \alpha =0$$

Applying double-angle formulas $\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$ and $\sin 2\alpha =2\sin \alpha \cos \alpha$:

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

Hence, we obtain the relation:

$$\beta =\cos 2\alpha +\sin 2\alpha$$

For $\beta =1$, the equation becomes:

$$\sin 2\alpha +\cos 2\alpha =1$$

To solve this trigonometric equation, we divide both sides by $\sqrt{2}$:

$$\frac{1}{\sqrt{2}}\sin 2\alpha +\frac{1}{\sqrt{2}}\cos 2\alpha =\frac{1}{\sqrt{2}}$$

Using the values $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ and $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$:

$$\cos 2\alpha \cos \frac{\pi }{4}+\sin 2\alpha \sin \frac{\pi }{4}=\cos \frac{\pi }{4}$$

By applying the trigonometric identity $\cos (A-B)=\cos A\cos B+\sin A\sin B$:

$$\cos \left(2\alpha -\frac{\pi }{4}\right)=\cos \frac{\pi }{4}$$

The general solution for $\cos \theta =\cos \varphi$ is $\theta =2n\pi \mp \varphi$ where $n\in \mathbb{Z}$. Therefore:

$$2\alpha -\frac{\pi }{4}=2n\pi \mp \frac{\pi }{4}$$

Solving for $2\alpha$ by adding $\frac{\pi }{4}$ to both sides:

$$2\alpha =2n\pi \mp \frac{\pi }{4}+\frac{\pi }{4}$$

This matches the correct general solution format:

$$2\alpha =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{4}$$