What is the general solution for 4+sin(a)=3sin^2(a) ?

The general solution for 4+sin(a)=3sin^2(a) is a=(3pi)/2+2pin

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4+sin(a)=3sin^2(a)

Solution

4 + sin (a) = 3 sin^{2} (a)

Solution

a = \frac{3 π}{2} + 2 πn

Degrees

a = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

4 + sin (a) = 3 sin^{2} (a)

Solve by substitution

4 + sin (a) = 3 sin^{2} (a)

Let:

sin (a) = u

4 + u = 3 u^{2}

4 + u = 3 u^{2} : u = \frac{4}{3}, u = - 1

4 + u = 3 u^{2}

Switch sides

3 u^{2} = 4 + u

Move

u

to the left side

3 u^{2} = 4 + u

Subtract

u

from both sides

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

Simplify

3 u^{2} - u = 4

3 u^{2} - u = 4

Move

4

to the left side

3 u^{2} - u = 4

Subtract

4

from both sides

3 u^{2} - u - 4 = 4 - 4

Simplify

3 u^{2} - u - 4 = 0

3 u^{2} - u - 4 = 0

Solve with the quadratic formula

3 u^{2} - u - 4 = 0

Quadratic Equation Formula:

For

a = 3, b = - 1, c = - 4

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

(- 1)^{2} - 4 \cdot 3 (- 4) = 7

(- 1)^{2} - 4 \cdot 3 (- 4)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 3 \cdot 4

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 3 \cdot 4 = 48

4 \cdot 3 \cdot 4

Multiply the numbers:

4 \cdot 3 \cdot 4 = 48

= 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 3}, u_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 3}

u = \frac{- ( - 1 ) + 7}{2 \cdot 3} : \frac{4}{3}

\frac{- ( - 1 ) + 7}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{1 + 7}{2 \cdot 3}

Add the numbers:

1 + 7 = 8

= \frac{8}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{8}{6}

Cancel the common factor:

2

= \frac{4}{3}

u = \frac{- ( - 1 ) - 7}{2 \cdot 3} : - 1

\frac{- ( - 1 ) - 7}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{1 - 7}{2 \cdot 3}

Subtract the numbers:

1 - 7 = - 6

= \frac{- 6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6}{6}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{4}{3}, u = - 1

Substitute back

u = sin (a)

sin (a) = \frac{4}{3}, sin (a) = - 1

sin (a) = \frac{4}{3}, sin (a) = - 1

sin (a) = \frac{4}{3} :

No Solution

sin (a) = \frac{4}{3}

- 1 \leq sin (x) \leq 1

No Solution

sin (a) = - 1 : a = \frac{3 π}{2} + 2 πn

sin (a) = - 1

General solutions for

sin (a) = - 1

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}

a = \frac{3 π}{2} + 2 πn

a = \frac{3 π}{2} + 2 πn

Combine all the solutions

a = \frac{3 π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4+sin(a)=3sin^2(a) ?
The general solution for 4+sin(a)=3sin^2(a) is a=(3pi)/2+2pin