What is the general solution for picos(2θ)=0 ?

The general solution for picos(2θ)=0 is θ= pi/4+pin,θ=(3pi)/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

picos(2θ)=0

Solution

π cos (2 θ) = 0

Solution

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

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

π cos (2 θ) = 0

Divide both sides by

π

π cos (2 θ) = 0

Divide both sides by

π

\frac{π cos ( 2 θ )}{π} = \frac{0}{π}

Simplify

cos (2 θ) = 0

cos (2 θ) = 0

General solutions for

cos (2 θ) = 0

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

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

Solve

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

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Solve

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

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

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

Popular Examples

2+sqrt(3)sec(x)-4cos(x)=2sqrt(3)

2 + 3 sec (x) - 4 cos (x) = 23

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

4 sin (x) \cdot cos (x) = 2 sin (x)

-1sec^2(θ)+2sec(θ)=1

- 1 sec^{2} (θ) + 2 sec (θ) = 1

cos(θ)=0.8701

cos (θ) = 0.8701

cos(3v)=cos(6),0<= v<= 360

cos (3 v) = cos (6), 0^{\circ} \leq v^{\circ} \leq 36 0^{\circ}

Frequently Asked Questions (FAQ)

What is the general solution for picos(2θ)=0 ?
The general solution for picos(2θ)=0 is θ= pi/4+pin,θ=(3pi)/4+pin