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

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

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

2 u^{2} = 3 - u

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

2 u^{2} = 3 - u

Move

u

to the left side

2 u^{2} = 3 - u

Add

u

to both sides

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

Simplify

2 u^{2} + u = 3

2 u^{2} + u = 3

Move

3

to the left side

2 u^{2} + u = 3

Subtract

3

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = 1, c = - 3

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 3)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 1 + 5 = 4

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Subtract the numbers:

- 1 - 5 = - 6

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 6}{4}

Apply the fraction rule:

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

= - \frac{6}{4}

Cancel the common factor:

2

= - \frac{3}{2}

The solutions to the quadratic equation are:

u = 1, u = - \frac{3}{2}

Substitute back

u = sin (x)

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

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

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) = - \frac{3}{2} :

No Solution

sin (x) = - \frac{3}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

Popular Examples

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9 tan (2 x) - 18 cos (x) = 0

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8 cos (2 t) - 14 sin (2 t) = 0

1-sqrt(2)sin(θ)=0

1 - 2 sin (θ) = 0

sin^2(x)=2cos(x)

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

2cos^2(θ)+4sin(θ)-13=9sin(θ)-9

2 cos^{2} (θ) + 4 sin (θ) - 13 = 9 sin (θ) - 9

Frequently Asked Questions (FAQ)

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