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

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

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

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

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

- 1 + tan (x) = 0

- 1 + tan (x) = 0

Move

1

to the right side

- 1 + tan (x) = 0

Add

1

to both sides

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

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

2 cos (x) = cot (x)

tan (x) = \frac{60}{\frac{( - 20 - 90 )}{2}}

0.1 \cdot sin (θ) cos (53.1 3^{\circ}) + 0.1 \cdot sin (θ) = 0

tan (θ) = (1)

sec^{2} (x) = \frac{3 3}{π}

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