Question

"Solve for all values of x on the given intervals. Write all answer in radians." I am stuck on number 4

Answer 1

`x=(2\pi)/(3)+2\pi n,x=(4\pi)/(3)+2\pi n`

Explanation:

Given the equation:

`\sin x\tan x=-2-\cot x\sin x`

Add 2+cot(x)sin(x) to both sides of the equation.

`\begin{gathered} \sin x\tan x+2+\cot x\sin x=-2-\cot x\sin x+2+\cot x\sin x \\ \sin x\tan x+2+\cot x\sin x=0 \end{gathered}`

Next, express in terms of sin and cos:

`\begin{gathered} \sin x(\sin x)/(\cos x)+2+(\cos x\sin x)/(\sin x)=0 \\ (\sin^2x)/(\cos x)+2+\cos x=0 \\ (\sin^2x+2\cos x+\cos^2x)/(\cos(x))=0 \\ \implies\sin^2x+2\cos x+\cos^2x=0 \end{gathered}`

Apply the Pythagorean Identity: cos²x+sinx=1

`2\cos x+1=0`

Subtract 1 from both sides:

`\begin{gathered} 2\cos x+1-1=0-1 \\ 2\cos x=-1 \end{gathered}`

Divide both sides by 2:

`\cos x=-(1)/(2)`

Take the arccos in the interval (-∞, ):

`\begin{gathered} x=\arccos(-0.5) \\ x=(2\pi)/(3)+2\pi n,x=(4\pi)/(3)+2\pi n \end{gathered}`

The values of x in the given interval are:

`x=(2\pi)/(3)+2\pi n,x=(4\pi)/(3)+2\pi n`