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= pi/2 ,x=pi

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

sin (x) - cos (x) = 1, 0 \leq x < 2 π

x = \frac{π}{2}, x = π

Solution steps

sin (x) - cos (x) = 1, 0 \leq x < 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{π}{4} + 2 πn, x - \frac{π}{4} = \frac{3 π}{4} + 2 πn

Solve

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

Solve

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

x = \frac{π}{2} + 2 πn, x = 2 πn + π

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{2}, x = π

tan (θ) = \frac{π}{6}

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

3 ∣ sin (3 θ) ∣ = 0

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

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

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= pi/2 ,x=pi