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

The general solution for 2cos(θ)+sqrt(2)=0,0<= θ<= 2pi is θ=(3pi)/4 ,θ=(5pi)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos(θ)+sqrt(2)=0,0<= θ<= 2pi

2 cos (θ) + 2 = 0, 0 \leq θ \leq 2 π

θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

Degrees

θ = 13 5^{\circ}, θ = 22 5^{\circ}

Solution steps

2 cos (θ) + 2 = 0, 0 \leq θ \leq 2 π

Move

2

to the right side

2 cos (θ) + 2 = 0

Subtract

2

from both sides

2 cos (θ) + 2 - 2 = 0 - 2

Simplify

2 cos (θ) = - 2

2 cos (θ) = - 2

Divide both sides by

2

2 cos (θ) = - 2

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 2}{2}

Simplify

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

General solutions for

cos (θ) = - \frac{2}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{3 π}{4}, θ = \frac{5 π}{4}

3 tan (\frac{θ}{2}) - 1 = 0

tan (\frac{x}{2}) = 3, 0 \leq x < 2 π

sin (x) = \frac{4}{cos ( x )}

tan^{3} (x) - 2 tan (x) = tan (x)

sin^{2} (x) + 2 cos^{2} (x) = 0

What is the general solution for 2cos(θ)+sqrt(2)=0,0<= θ<= 2pi ?
The general solution for 2cos(θ)+sqrt(2)=0,0<= θ<= 2pi is θ=(3pi)/4 ,θ=(5pi)/4