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

The general solution for 1+4sin^2(θ)=2 is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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1+4sin^2(θ)=2

Solution

1 + 4 sin^{2} (θ) = 2

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n, θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

1 + 4 sin^{2} (θ) = 2

Solve by substitution

1 + 4 sin^{2} (θ) = 2

Let:

sin (θ) = u

1 + 4 u^{2} = 2

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

1 + 4 u^{2} = 2

Move

1

to the right side

1 + 4 u^{2} = 2

Subtract

1

from both sides

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

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

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

- \frac{1}{4}

Simplify

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

\frac{1}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

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

Substitute back

u = sin (θ)

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

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

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 1+4sin^2(θ)=2 ?
The general solution for 1+4sin^2(θ)=2 is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin