Question

Using the properties of determinants solve for x.
12 Grade Math

Answer 1

Answer:

    nπ   or   π/6 + 2nπ   or   5π/6 + 2nπ

Step-by-step explanation:

Expanding the determinant along the third column gives:

$\displaystyle\left|\begin{matrix}1&1&\sin3x\\-4&3&\cos2x\\7&-7&-2\end{matrix}\right|=0\\\\\Rightarrow\left|\begin{matrix}-4&3\\7&-7\end{matrix}\right|\sin3x-\left|\begin{matrix}1&1\\7&-7\end{matrix}\right|\cos2x+\left|\begin{matrix}1&1\\-4&3\end{matrix}\right|\times(-2)=0\\\\\Rightarrow7\sin3x+14\cos2x-14=0\\\\\Rightarrow\sin3x+2\cos2x-2=0$

Now express just in terms of sin x by using:

   cos 2x = 1 - 2 sin²x

and

   sin 3x = sin (2x + x)

              = sin 2x cos x + cos 2x sin x

              = 2 sin x cos²x + ( 1 - 2 sin²x ) sin x

              = 2 sin x ( 1 - sin²x ) + sin x - 2 sin³ x

              = 3 sin x  -  4 sin³x

So the equation above becomes:

$\displaystyle3\sin x-4\sin^3x+2-4\sin^2x-2=0\\\Rightarrow\sin x(4\sin^2 x+4\sin x-3)=0\\\Rightarrow\sin x(2\sin x-1)(2\sin x+3)=0\\\Rightarrow\sin x(2\sin x-1)=0\quad\quad\text{(since 2\sin x+3\neq0 )}\\\Rightarrow\sin x=0\quad\text{or}\quad\sin x=\tfrac12\\\Rightarrow x=n\pi\quad\text{or}\quad x=\frac{\pi}6+2n\pi\quad\text{or}\quad x=\frac{5\pi}6+2n\pi$