What is the general solution for (-cos(x)+sin(x))/(1+sin(2x))=0 ?

The general solution for (-cos(x)+sin(x))/(1+sin(2x))=0 is x= pi/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\frac{- cos ( x ) + sin ( x )}{1 + sin ( 2 x )} = 0

x = \frac{π}{4} + πn

Solution steps

\frac{- cos ( x ) + sin ( x )}{1 + sin ( 2 x )} = 0

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

- cos (x) + sin (x) = 0

Rewrite using trig identities

- 1 + tan (x) = 0

Move

1

to the right side

tan (x) = 1

General solutions for

tan (x) = 1

x = \frac{π}{4} + πn

cos^{2} (2 x) - \frac{1}{4} = 0

cosh (x) = \frac{17}{15}

4 cos (x) + 23 = 0

c, 1 = tan (c)

4 cos^{2} (x) + 3 sin^{2} (x) - 4 = - tan^{2} (x)

What is the general solution for (-cos(x)+sin(x))/(1+sin(2x))=0 ?
The general solution for (-cos(x)+sin(x))/(1+sin(2x))=0 is x= pi/4+pin