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

Q: 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

Solution

sin (x) + cos (x) = 1, 0 \leq x < 2 π

Solution

x = 0, x = \frac{π}{2}

Degrees

x = 0^{\circ}, x = 9 0^{\circ}

Solution steps

sin (x) + cos (x) = 1, 0 \leq x < 2 π

Rewrite using trig identities

sin (x) + cos (x)

sin (x) + cos (x) = 2 sin (x + \frac{π}{4})

sin (x) + cos (x)

Rewrite as

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (x) + sin (\frac{π}{4}) cos (x))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (x + \frac{π}{4})

= 2 sin (x + \frac{π}{4})

2 sin (x + \frac{π}{4}) = 1

Divide both sides by

2

2 sin (x + \frac{π}{4}) = 1

Divide both sides by

2

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{1}{2}

Simplify

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{1}{2}

Simplify

\frac{2 sin ( x + \frac{π}{4} )}{2} : sin (x + \frac{π}{4})

\frac{2 sin ( x + \frac{π}{4} )}{2}

Cancel the common factor:

2

= sin (x + \frac{π}{4})

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

sin (x + \frac{π}{4}) = \frac{2}{2}

sin (x + \frac{π}{4}) = \frac{2}{2}

sin (x + \frac{π}{4}) = \frac{2}{2}

General solutions for

sin (x + \frac{π}{4}) = \frac{2}{2}

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

x + \frac{π}{4} = \frac{π}{4} + 2 πn

Subtract

\frac{π}{4}

from both sides

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

Simplify

x = 2 πn

Solve

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

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

Move

\frac{π}{4}

to the right side

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

Subtract

\frac{π}{4}

from both sides

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

Simplify

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

Simplify

x + \frac{π}{4} - \frac{π}{4} : x

x + \frac{π}{4} - \frac{π}{4}

Add similar elements:

\frac{π}{4} - \frac{π}{4} = 0

= x

Simplify

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

\frac{3 π}{4} + 2 πn - \frac{π}{4}

Group like terms

= 2 πn - \frac{π}{4} + \frac{3 π}{4}

Combine the fractions

- \frac{π}{4} + \frac{3 π}{4} : \frac{π}{2}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- π + 3 π}{4}

Add similar elements:

- π + 3 π = 2 π

= \frac{2 π}{4}

Cancel the common factor:

2

= \frac{π}{2}

= 2 πn + \frac{π}{2}

x = 2 πn + \frac{π}{2}

x = 2 πn + \frac{π}{2}

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}

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Frequently Asked Questions (FAQ)

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