What is the general solution for 2sin^2(x)=4sin(x)+6,0<= x<= 2pi ?

Q: What is the general solution for 2sin^2(x)=4sin(x)+6,0<= x<= 2pi ?

The general solution for 2sin^2(x)=4sin(x)+6,0<= x<= 2pi is x=(3pi)/2

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2sin^2(x)=4sin(x)+6,0<= x<= 2pi

Solution

2 sin^{2} (x) = 4 sin (x) + 6, 0 \leq x \leq 2 π

Solution

x = \frac{3 π}{2}

Degrees

x = 27 0^{\circ}

Solution steps

2 sin^{2} (x) = 4 sin (x) + 6, 0 \leq x \leq 2 π

Solve by substitution

2 sin^{2} (x) = 4 sin (x) + 6

Let:

sin (x) = u

2 u^{2} = 4 u + 6

2 u^{2} = 4 u + 6 : u = 3, u = - 1

2 u^{2} = 4 u + 6

Move

6

to the left side

2 u^{2} = 4 u + 6

Subtract

6

from both sides

2 u^{2} - 6 = 4 u + 6 - 6

Simplify

2 u^{2} - 6 = 4 u

2 u^{2} - 6 = 4 u

Move

4 u

to the left side

2 u^{2} - 6 = 4 u

Subtract

4 u

from both sides

2 u^{2} - 6 - 4 u = 4 u - 4 u

Simplify

2 u^{2} - 6 - 4 u = 0

2 u^{2} - 6 - 4 u = 0

Write in the standard form

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

2 u^{2} - 4 u - 6 = 0

Solve with the quadratic formula

2 u^{2} - 4 u - 6 = 0

Quadratic Equation Formula:

For

a = 2, b = - 4, c = - 6

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 (- 6) = 8

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} + 4 \cdot 2 \cdot 6

Multiply the numbers:

4 \cdot 2 \cdot 6 = 48

= 4^{2} + 48

4^{2} = 16

= 16 + 48

Add the numbers:

16 + 48 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 2}, u_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 2}

u = \frac{- ( - 4 ) + 8}{2 \cdot 2} : 3

\frac{- ( - 4 ) + 8}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{4 + 8}{2 \cdot 2}

Add the numbers:

4 + 8 = 12

= \frac{12}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= 3

u = \frac{- ( - 4 ) - 8}{2 \cdot 2} : - 1

\frac{- ( - 4 ) - 8}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{4 - 8}{2 \cdot 2}

Subtract the numbers:

4 - 8 = - 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 = 3, u = - 1

Substitute back

u = sin (x)

sin (x) = 3, sin (x) = - 1

sin (x) = 3, sin (x) = - 1

sin (x) = 3, 0 \leq x \leq 2 π :

No Solution

sin (x) = 3, 0 \leq x \leq 2 π

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{2}

sin (x) = - 1, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{3 π}{2}

Combine all the solutions

x = \frac{3 π}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)=4sin(x)+6,0<= x<= 2pi ?
The general solution for 2sin^2(x)=4sin(x)+6,0<= x<= 2pi is x=(3pi)/2