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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(θ)-cos(θ)=0,0<= θ<= 2pi

sin (θ) - cos (θ) = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{4}, θ = \frac{5 π}{4}

Degrees

θ = 4 5^{\circ}, θ = 22 5^{\circ}

Solution steps

sin (θ) - cos (θ) = 0, 0 \leq θ \leq 2 π

Rewrite using trig identities

sin (θ) - cos (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{sin ( θ ) - cos ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

\frac{sin ( θ )}{cos ( θ )} - 1 = 0

Use the basic trigonometric identity:

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

tan (θ) - 1 = 0

tan (θ) - 1 = 0

Move

1

to the right side

tan (θ) - 1 = 0

Add

1

to both sides

tan (θ) - 1 + 1 = 0 + 1

Simplify

tan (θ) = 1

tan (θ) = 1

General solutions for

tan (θ) = 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}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{4}, θ = \frac{5 π}{4}

3 sec^{2} (θ) - 12 = 5 sec (θ)

3 cos (x) - 3 = cos (x)

sin (x) - 1 = - cos (x)

36 cos (2 x) = 0

2 sin^{2} (θ) + sin (θ) = 0, 0 \leq θ < 2 π

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