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

The general solution for 0=a(1+cos(2θ)) is θ= pi/2+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0=a(1+cos(2θ))

0 = a (1 + cos (2 θ))

θ = \frac{π}{2} + πn

Degrees

θ = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

0 = a (1 + cos (2 θ))

Switch sides

a (1 + cos (2 θ)) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

1 + cos (2 θ) = 0

Move

1

to the right side

1 + cos (2 θ) = 0

Subtract

1

from both sides

1 + cos (2 θ) - 1 = 0 - 1

Simplify

cos (2 θ) = - 1

cos (2 θ) = - 1

General solutions for

cos (2 θ) = - 1

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}

2 θ = π + 2 πn

2 θ = π + 2 πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

2 θ = π + 2 πn

Divide both sides by

2

2 θ = π + 2 πn

Divide both sides by

2

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

Simplify

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

4 sin (2 x) = 3 cos (x)

sin (b) = (\frac{250 km}{590 km}) sin (2 3^{\circ}), 8 2^{\circ}

sin (x) = \frac{30}{60}

4 cos (z) + 5 = 0

y, sin (x^{2} y^{2}) = x

What is the general solution for 0=a(1+cos(2θ)) ?
The general solution for 0=a(1+cos(2θ)) is θ= pi/2+pin