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 ,x=2pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = 0, x = \frac{π}{2}, x = 2 π

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{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 \leq 2 π

x = 0, x = \frac{π}{2}, x = 2 π

1 sin (x) tan (x) - 3 sin (x) = 0

sec (A) [1 - sin^{2} (A)] = 1

3 sin^{2} (x) - 8 sin (x) - 3 = 0

6 sin (x) - 6 = 0

(cot (x) - 1) (csc (x) + 1) = 0

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 ,x=2pi