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.

Step-by-step explanation:

It is given that,

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

Then,

`\begin{gathered} 8*2(\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(√(2)\sin x+1)(√(2)sinx-1)=0 \\ 8\cos x=0,√(2)\sin x+1=0,√(2)\sin x-1=0 \\ \cos x=0,\sin x=-(1)/(√(2)),\sin x=(1)/(√(2)) \end{gathered}`

That implies,

`\begin{gathered} x=\cos^(-1)(0),x=\sin^(-1)((-1)/(√(2))),x=\sin^(-1)((1)/(√(2))) \\ x=((\pi)/(2),(3\pi)/(2)),x=((5\pi)/(4),(7\pi)/(4)),x=((\pi)/(4),(3\pi)/(4)) \end{gathered}`

Hence, the answeris,

`x=(\pi)/(2),(3\pi)/(2),(5\pi)/(4),(7\pi)/(4),(\pi)/(4),(3\pi)/(4)`