What is the general solution for tan(x)+sin(x)tan(x)=0,-pi<= x<pi ?

The general solution for tan(x)+sin(x)tan(x)=0,-pi<= x<pi is x=0,x=-pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan(x)+sin(x)tan(x)=0,-pi<= x<pi

tan (x) + sin (x) tan (x) = 0, - π \leq x < π

x = 0, x = - π

Solution steps

tan (x) + sin (x) tan (x) = 0, - π \leq x < π

Factor

tan (x) + sin (x) tan (x) : tan (x) (sin (x) + 1)

tan (x) (sin (x) + 1) = 0

Solving each part separately

tan (x) = 0 or sin (x) + 1 = 0

tan (x) = 0, - π \leq x < π : x = 0, x = - π

sin (x) + 1 = 0, - π \leq x < π : x = - \frac{π}{2}

Combine all the solutions

x = 0, x = - π, x = - \frac{π}{2}

Since the equation is undefined for:

- \frac{π}{2}

x = 0, x = - π

4 sin^{2} (x) - sin (x) - 2 = 0

2 cos (x) sin (x) + 1 cos (x) = 0

sin^{2} (x) = - 2 cos^{2} (x)

3 sin (x) = - 2

6 sin (x) + 1 = 4 sin (x)

What is the general solution for tan(x)+sin(x)tan(x)=0,-pi<= x<pi ?
The general solution for tan(x)+sin(x)tan(x)=0,-pi<= x<pi is x=0,x=-pi