What is the general solution for 2cos(2θ)+1=0,0<= θ<= 2pi ?

The general solution for 2cos(2θ)+1=0,0<= θ<= 2pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(5pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos(2θ)+1=0,0<= θ<= 2pi

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

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

Solution steps

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

Move

1

to the right side

2 cos (2 θ) = - 1

Divide both sides by

2

cos (2 θ) = - \frac{1}{2}

General solutions for

cos (2 θ) = - \frac{1}{2}

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

Solve

2 θ = \frac{2 π}{3} + 2 πn : θ = \frac{π}{3} + πn

Solve

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

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

Solutions for the range

0 \leq θ \leq 2 π

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

cot (x) sin (x) + cot (x) = 0

4 cos (2 θ) = 4 sin (2 θ)

tan (x) = - 3, 0 \leq x \leq 2 π

5 tan^{2} (A) - 7 = 8

tan (x) = \frac{28}{16}

What is the general solution for 2cos(2θ)+1=0,0<= θ<= 2pi ?
The general solution for 2cos(2θ)+1=0,0<= θ<= 2pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(5pi)/3