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

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

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sin^2(x)=-3sin(x)+4

Solution

sin^{2} (x) = - 3 sin (x) + 4

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (x) = - 3 sin (x) + 4

Solve by substitution

sin^{2} (x) = - 3 sin (x) + 4

Let:

sin (x) = u

u^{2} = - 3 u + 4

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

u^{2} = - 3 u + 4

Move

4

to the left side

u^{2} = - 3 u + 4

Subtract

4

from both sides

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

Simplify

u^{2} - 4 = - 3 u

u^{2} - 4 = - 3 u

Move

3 u

to the left side

u^{2} - 4 = - 3 u

Add

3 u

to both sides

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

Simplify

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

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

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

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 1 \cdot 4

Multiply the numbers:

4 \cdot 1 \cdot 4 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 3 + 5}{2 \cdot 1}, u_{2} = \frac{- 3 - 5}{2 \cdot 1}

u = \frac{- 3 + 5}{2 \cdot 1} : 1

\frac{- 3 + 5}{2 \cdot 1}

Add/Subtract the numbers:

- 3 + 5 = 2

= \frac{2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 3 - 5}{2 \cdot 1} : - 4

\frac{- 3 - 5}{2 \cdot 1}

Subtract the numbers:

- 3 - 5 = - 8

= \frac{- 8}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 8}{2}

Apply the fraction rule:

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

= - \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= - 4

The solutions to the quadratic equation are:

u = 1, u = - 4

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - 4

sin (x) = 1, sin (x) = - 4

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = - 4 :

No Solution

sin (x) = - 4

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

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