What is the general solution for sin(x)-cos(x)=-1,0<= x<= 2pi ?

The general solution for sin(x)-cos(x)=-1,0<= x<= 2pi is x=(3pi)/2 ,x=2pi,x=0

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sin(x)-cos(x)=-1,0<= x<= 2pi

sin (x) - cos (x) = - 1, 0 \leq x \leq 2 π

x = \frac{3 π}{2}, x = 2 π, x = 0

Solution steps

sin (x) - cos (x) = - 1, 0 \leq x \leq 2 π

Rewrite using trig identities

2 sin (x - \frac{π}{4}) = - 1

Divide both sides by

2

sin (x - \frac{π}{4}) = - \frac{2}{2}

General solutions for

sin (x - \frac{π}{4}) = - \frac{2}{2}

x - \frac{π}{4} = \frac{5 π}{4} + 2 πn, x - \frac{π}{4} = \frac{7 π}{4} + 2 πn

Solve

x - \frac{π}{4} = \frac{5 π}{4} + 2 πn : x = 2 πn + \frac{3 π}{2}

Solve

x - \frac{π}{4} = \frac{7 π}{4} + 2 πn : x = 2 πn + 2 π

x = 2 πn + \frac{3 π}{2}, x = 2 πn + 2 π

Solutions for the range

0 \leq x \leq 2 π

x = \frac{3 π}{2}, x = 2 π, x = 0

cos (θ) = (\frac{1}{2})

sin^{2} (x) - cos^{2} (x) + sin (x) = 0, 0 \leq x \leq 2 π

cos (θ) = - cos (2 θ)

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

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

What is the general solution for sin(x)-cos(x)=-1,0<= x<= 2pi ?
The general solution for sin(x)-cos(x)=-1,0<= x<= 2pi is x=(3pi)/2 ,x=2pi,x=0