What is the general solution for cos(2θ)(4sin^2(θ)+1)=0 ?

The general solution for cos(2θ)(4sin^2(θ)+1)=0 is θ= pi/4+pin,θ=(3pi)/4+pin

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cos(2θ)(4sin^2(θ)+1)=0

Solution

cos (2 θ) (4 sin^{2} (θ) + 1) = 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

cos (2 θ) (4 sin^{2} (θ) + 1) = 0

Solving each part separately

cos (2 θ) = 0 or 4 sin^{2} (θ) + 1 = 0

cos (2 θ) = 0 : θ = \frac{π}{4} + πn, θ = \frac{3 π}{4} + πn

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

4 sin^{2} (θ) + 1 = 0 :

No Solution

4 sin^{2} (θ) + 1 = 0

Solve by substitution

4 sin^{2} (θ) + 1 = 0

Let:

sin (θ) = u

4 u^{2} + 1 = 0

4 u^{2} + 1 = 0 : u = i \frac{1}{2}, u = - i \frac{1}{2}

4 u^{2} + 1 = 0

Move

1

to the right side

4 u^{2} + 1 = 0

Subtract

1

from both sides

4 u^{2} + 1 - 1 = 0 - 1

Simplify

4 u^{2} = - 1

4 u^{2} = - 1

Divide both sides by

4

4 u^{2} = - 1

Divide both sides by

4

\frac{4 u ^{2}}{4} = \frac{- 1}{4}

Simplify

u^{2} = - \frac{1}{4}

u^{2} = - \frac{1}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - \frac{1}{4}, u = - - \frac{1}{4}

Simplify

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

- a = - 1 a

- \frac{1}{4} = - 1 \frac{1}{4}

= - 1 \frac{1}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= i \frac{1}{2}

Apply rule

1 = 1

= i \frac{1}{2}

Rewrite

i \frac{1}{2}

in standard complex form:

\frac{1}{2} i

i \frac{1}{2}

Multiply fractions:

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

= \frac{1 i}{2}

Multiply:

1 i = i

= \frac{i}{2}

= \frac{1}{2} i

Simplify

- - \frac{1}{4} : - i \frac{1}{2}

- - \frac{1}{4}

Simplify

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

- a = - 1 a

- \frac{1}{4} = - 1 \frac{1}{4}

= - 1 \frac{1}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= i \frac{1}{2}

= - i \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2} i

u = i \frac{1}{2}, u = - i \frac{1}{2}

Substitute back

u = sin (θ)

sin (θ) = i \frac{1}{2}, sin (θ) = - i \frac{1}{2}

sin (θ) = i \frac{1}{2}, sin (θ) = - i \frac{1}{2}

sin (θ) = i \frac{1}{2} :

No Solution

sin (θ) = i \frac{1}{2}

No Solution

sin (θ) = - i \frac{1}{2} :

No Solution

sin (θ) = - i \frac{1}{2}

No Solution

Combine all the solutions

No Solution

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

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