What is the general solution for (sin(x))/(sin(2x))=0 ?

The general solution for (sin(x))/(sin(2x))=0 is No Solution for x\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(sin(x))/(sin(2x))=0

\frac{sin ( x )}{sin ( 2 x )} = 0

No Solution for x \in R

Solution steps

\frac{sin ( x )}{sin ( 2 x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Since the equation is undefined for:

2 πn, π + 2 πn

No Solution for x \in R

5 cos^{2} (x) = \frac{1}{2 ( 1 + cos ^{2} ( x ) )}

2 sin^{2} (\frac{x}{2}) - 1 = sin (\frac{x}{2})

4.7 2^{2} = 3. 3^{2} + 3. 3^{2} - 2 (3.3) (3.3) cos (x)

2 sin (x - 3 0^{\circ}) = cos (x - 6 0^{\circ})

1 = 2 cos (3 x)

What is the general solution for (sin(x))/(sin(2x))=0 ?
The general solution for (sin(x))/(sin(2x))=0 is No Solution for x\in\mathbb{R}