Answer:
[tex]x=\frac{2\pi}{3}+2\pi n,x=\frac{4\pi}{3}+2\pi n[/tex]
Explanation:
Given the equation:
[tex]\sin x\tan x=-2-\cot x\sin x[/tex]
Add 2+cot(x)sin(x) to both sides of the equation.
[tex]\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}[/tex]
Next, express in terms of sin and cos:
[tex]\begin{gathered} \sin x\frac{\sin x}{\cos x}+2+\frac{\cos x\sin x}{\sin x}=0 \\ \frac{\sin^2x}{\cos x}+2+\cos x=0 \\ \frac{\sin^2x+2\cos x+\cos^2x}{\cos(x)}=0 \\ \implies\sin^2x+2\cos x+\cos^2x=0 \end{gathered}[/tex]
Apply the Pythagorean Identity: cos²x+sinx=1
[tex]2\cos x+1=0[/tex]
Subtract 1 from both sides:
[tex]\begin{gathered} 2\cos x+1-1=0-1 \\ 2\cos x=-1 \end{gathered}[/tex]
Divide both sides by 2:
[tex]\cos x=-\frac{1}{2}[/tex]
Take the arccos in the interval (-∞, ):
[tex]\begin{gathered} x=\arccos(-0.5) \\ x=\frac{2\pi}{3}+2\pi n,x=\frac{4\pi}{3}+2\pi n \end{gathered}[/tex]
The values of x in the given interval are:
[tex]x=\frac{2\pi}{3}+2\pi n,x=\frac{4\pi}{3}+2\pi n[/tex]
