What is the general solution for 2sin^2(θ)=2+cos(2θ) ?

The general solution for 2sin^2(θ)=2+cos(2θ) is θ= pi/3+pin,θ=(2pi)/3+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(θ)=2+cos(2θ)

Solution

2 sin^{2} (θ) = 2 + cos (2 θ)

Solution

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

Degrees

θ = 6 0^{\circ} + 18 0^{\circ} n, θ = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

2 sin^{2} (θ) = 2 + cos (2 θ)

Subtract

2 + cos (2 θ)

from both sides

2 sin^{2} (θ) - 2 - cos (2 θ) = 0

Rewrite using trig identities

- 2 - cos (2 θ) + 2 sin^{2} (θ)

Use the Double Angle identity:

1 - 2 sin^{2} (x) = cos (2 x)

- 2 sin^{2} (x) = cos (2 x) - 1

= - 2 - cos (2 θ) - (cos (2 θ) - 1)

Simplify

- 2 - cos (2 θ) - (cos (2 θ) - 1) : - 2 cos (2 θ) - 1

- 2 - cos (2 θ) - (cos (2 θ) - 1)

- (cos (2 θ) - 1) : - cos (2 θ) + 1

- (cos (2 θ) - 1)

Distribute parentheses

= - (cos (2 θ)) - (- 1)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - cos (2 θ) + 1

= - 2 - cos (2 θ) - cos (2 θ) + 1

Simplify

- 2 - cos (2 θ) - cos (2 θ) + 1 : - 2 cos (2 θ) - 1

- 2 - cos (2 θ) - cos (2 θ) + 1

Add similar elements:

- cos (2 θ) - cos (2 θ) = - 2 cos (2 θ)

= - 2 - 2 cos (2 θ) + 1

Group like terms

= - 2 cos (2 θ) - 2 + 1

Add/Subtract the numbers:

- 2 + 1 = - 1

= - 2 cos (2 θ) - 1

= - 2 cos (2 θ) - 1

= - 2 cos (2 θ) - 1

- 1 - 2 cos (2 θ) = 0

Move

1

to the right side

- 1 - 2 cos (2 θ) = 0

Add

1

to both sides

- 1 - 2 cos (2 θ) + 1 = 0 + 1

Simplify

- 2 cos (2 θ) = 1

- 2 cos (2 θ) = 1

Divide both sides by

- 2

- 2 cos (2 θ) = 1

Divide both sides by

- 2

\frac{- 2 cos ( 2 θ )}{- 2} = \frac{1}{- 2}

Simplify

cos (2 θ) = - \frac{1}{2}

cos (2 θ) = - \frac{1}{2}

General solutions for

cos (2 θ) = - \frac{1}{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}

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

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

2 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

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

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 π}{6}

Cancel the common factor:

2

= \frac{π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

Solve

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

2 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

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

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

\frac{\frac{4 π}{3}}{2}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{4 π}{6}

Cancel the common factor:

2

= \frac{2 π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

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

Popular Examples

tan^2(x)-1=0,0<= x<= 2pi

tan^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

sin^2(θ)= 1/4

sin^{2} (θ) = \frac{1}{4}

(tan(θ)+1)(sec(θ)+1)=0

(tan (θ) + 1) (sec (θ) + 1) = 0

2sin(θ)-tan^2(θ)=0

2 sin (θ) - tan^{2} (θ) = 0

sin(θ)-sqrt(3)cos(θ)=1

sin (θ) - 3 cos (θ) = 1

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(θ)=2+cos(2θ) ?
The general solution for 2sin^2(θ)=2+cos(2θ) is θ= pi/3+pin,θ=(2pi)/3+pin