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

The general solution for cos(x)=sin(x),0<= x<= 2pi is x= pi/4 ,x=(5pi)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

cos (x) = sin (x), 0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{5 π}{4}

Degrees

x = 4 5^{\circ}, x = 22 5^{\circ}

Solution steps

cos (x) = sin (x), 0 \leq x \leq 2 π

Rewrite using trig identities

cos (x) = sin (x)

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{cos ( x )}{cos ( x )} = \frac{sin ( x )}{cos ( x )}

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 = tan (x)

Switch sides

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{5 π}{4}

sin (4 (x - \frac{π}{4})) = 0

3 arcsin (2 θ) = 2 π

cos^{3} (x) + cos^{2} (x) - cos (x) - 1 = 0

tan (2 θ) = 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

2 cos (θ) + 2 = 0, 0 \leq θ \leq 2 π

What is the general solution for cos(x)=sin(x),0<= x<= 2pi ?
The general solution for cos(x)=sin(x),0<= x<= 2pi is x= pi/4 ,x=(5pi)/4