Question

Solve the equation. (Find all solutions of the equation in the interval [0, 2). Enter your answers as a comma-separated list.)8 sin(2x) sin(x) = 8 cos(x)

Answer 1

Given

The equation,

$8\sin(2x)\sin(x)=8\cos(x)$

To solve for x.

Explanation:

It is given that,

$8\sin(2x)\sin(x)=8\cos(x)$

Then,

$\begin{gathered} 8\times2(\sin(x)\cos(x))\sin(x)-8\cos(x)=0 \\ 16\sin^2(x)\cos x-8\cos x=0 \\ 8\cos x(2\sin^2x-1)=0 \\ 8\cos x(\sqrt{2}\sin x+1)(\sqrt{2}sinx-1)=0 \\ 8\cos x=0,\sqrt{2}\sin x+1=0,\sqrt{2}\sin x-1=0 \\ \cos x=0,\sin x=-\frac{1}{\sqrt{2}},\sin x=\frac{1}{\sqrt{2}} \end{gathered}$

That implies,

$\begin{gathered} x=\cos^{-1}(0),x=\sin^{-1}(\frac{-1}{\sqrt{2}}),x=\sin^{-1}(\frac{1}{\sqrt{2}}) \\ x=(\frac{\pi}{2},\frac{3\pi}{2}),x=(\frac{5\pi}{4},\frac{7\pi}{4}),x=(\frac{\pi}{4},\frac{3\pi}{4}) \end{gathered}$

Hence, the answeris,

$x=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{4},\frac{7\pi}{4},\frac{\pi}{4},\frac{3\pi}{4}$