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

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(-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

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