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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

2 u^{2} = 2 u + 4

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

2 u^{2} = 2 u + 4

Move

4

to the left side

2 u^{2} = 2 u + 4

Subtract

4

from both sides

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

Simplify

2 u^{2} - 4 = 2 u

2 u^{2} - 4 = 2 u

Move

2 u

to the left side

2 u^{2} - 4 = 2 u

Subtract

2 u

from both sides

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

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 2, c = - 4

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

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

(- 2)^{2} - 4 \cdot 2 (- 4) = 6

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 2 \cdot 4 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

Add the numbers:

4 + 32 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 2}, u_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 2}

u = \frac{- ( - 2 ) + 6}{2 \cdot 2} : 2

\frac{- ( - 2 ) + 6}{2 \cdot 2}

Apply rule

- (- a) = a

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

Add the numbers:

2 + 6 = 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= 2

u = \frac{- ( - 2 ) - 6}{2 \cdot 2} : - 1

\frac{- ( - 2 ) - 6}{2 \cdot 2}

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 6 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 2, u = - 1

Substitute back

u = sin (x)

sin (x) = 2, sin (x) = - 1

sin (x) = 2, sin (x) = - 1

sin (x) = 2 :

No Solution

sin (x) = 2

- 1 \leq sin (x) \leq 1

No Solution

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

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Popular Examples

sin(x)cos^2(x)=sin(x),0<= x<= 2pi

sin (x) cos^{2} (x) = sin (x), 0 \leq x \leq 2 π

2*cos(4x)-1=0

2 \cdot cos (4 x) - 1 = 0

2cos^2(θ)-3cos(θ)=2

2 cos^{2} (θ) - 3 cos (θ) = 2

arctan(0.2x)=40

arctan (0.2 x) = 4 0^{\circ}

sin(x)-sin(2x)+sin(3x)=0

sin (x) - sin (2 x) + sin (3 x) = 0

Frequently Asked Questions (FAQ)

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