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

The general solution for sin(x)=cos(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 >

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

sin (x) = cos (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

sin (x) = cos (x), 0 \leq x \leq 2 π

Rewrite using trig identities

sin (x) = cos (x)

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

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}

3 cos (θ) = 0

2 sin^{2} (θ) + sin (θ) - 9 = 2 sin (θ) - 8

cos (θ) = \frac{5}{3}

- 16 sin (x) + 8 sin (2 x) = 0

csc (x) = - \frac{1}{2}

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