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=0,x= pi/2

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

sin (x) + cos (x) = 1, 0 \leq x < 2 π

x = 0, x = \frac{π}{2}

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 = 2 πn

Solve

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

x = 2 πn, x = 2 πn + \frac{π}{2}

Solutions for the range

0 \leq x < 2 π

x = 0, x = \frac{π}{2}

x, cos^{2} (x + y) = sin^{2} (x + y)

cot^{2} (x) + 3 cot (x) + 2 = cot (x) + 1

a^{2} = b^{2} + c^{2} - 2 b c \cdot cos (x)

3 sin^{2} (X) - cos (X) = 0

4 sin (θ) = csc (θ)

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=0,x= pi/2