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

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

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

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

x = \frac{π}{4} + πn

Degrees

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

Solution steps

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

Rewrite using trig identities

- 1 + cos (x) + cos^{2} (x) - sin (x) + sin^{2} (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= - 1 + cos (x) - sin (x) + 1

Simplify

- 1 + cos (x) - sin (x) + 1 : cos (x) - sin (x)

- 1 + cos (x) - sin (x) + 1

Group like terms

= cos (x) - sin (x) - 1 + 1

- 1 + 1 = 0

= cos (x) - sin (x)

= cos (x) - sin (x)

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

Subtract

1

from both sides

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

Simplify

- tan (x) = - 1

- tan (x) = - 1

Divide both sides by

- 1

- tan (x) = - 1

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 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

sin (4 x) - cos (2 x) = 0

\frac{6}{π} = 3 cos (2 x)

sin (2 x) = - 2 cos (x)

4 - 8 cos^{2} (x) = 0

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

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