What is the general solution for 2sin^2(x)+1=4sin(x) ?

The general solution for 2sin^2(x)+1=4sin(x) is x=0.29725…+2pin,x=pi-0.29725…+2pin

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2sin^2(x)+1=4sin(x)

Solution

2 sin^{2} (x) + 1 = 4 sin (x)

Solution

x = 0.29725 \dots + 2 πn, x = π - 0.29725 \dots + 2 πn

Degrees

x = 17.03124 \dots^{\circ} + 36 0^{\circ} n, x = 162.96875 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (x) + 1 = 4 sin (x)

Solve by substitution

2 sin^{2} (x) + 1 = 4 sin (x)

Let:

sin (x) = u

2 u^{2} + 1 = 4 u

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

2 u^{2} + 1 = 4 u

Move

4 u

to the left side

2 u^{2} + 1 = 4 u

Subtract

4 u

from both sides

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

Simplify

2 u^{2} + 1 - 4 u = 0

2 u^{2} + 1 - 4 u = 0

Write in the standard form

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

2 u^{2} - 4 u + 1 = 0

Solve with the quadratic formula

2 u^{2} - 4 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 4, c = 1

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

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

(- 4)^{2} - 4 \cdot 2 \cdot 1 = 22

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

Apply exponent rule:

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

n

is even

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

= 4^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 4^{2} - 8

4^{2} = 16

= 16 - 8

Subtract the numbers:

16 - 8 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 2 = 4

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

Factor

4 + 22 : 2 (2 + 2)

4 + 22

Rewrite as

= 2 \cdot 2 + 22

Factor out common term

2

= 2 (2 + 2)

= \frac{2 ( 2 + 2 )}{4}

Cancel the common factor:

2

= \frac{2 + 2}{2}

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 2 = 4

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

Factor

4 - 22 : 2 (2 - 2)

4 - 22

Rewrite as

= 2 \cdot 2 - 22

Factor out common term

2

= 2 (2 - 2)

= \frac{2 ( 2 - 2 )}{4}

Cancel the common factor:

2

= \frac{2 - 2}{2}

The solutions to the quadratic equation are:

u = \frac{2 + 2}{2}, u = \frac{2 - 2}{2}

Substitute back

u = sin (x)

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

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

sin (x) = \frac{2 + 2}{2} :

No Solution

sin (x) = \frac{2 + 2}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{2 - 2}{2} : x = arcsin (\frac{2 - 2}{2}) + 2 πn, x = π - arcsin (\frac{2 - 2}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{2 - 2}{2}) + 2 πn, x = π - arcsin (\frac{2 - 2}{2}) + 2 πn

x = arcsin (\frac{2 - 2}{2}) + 2 πn, x = π - arcsin (\frac{2 - 2}{2}) + 2 πn

Combine all the solutions

x = arcsin (\frac{2 - 2}{2}) + 2 πn, x = π - arcsin (\frac{2 - 2}{2}) + 2 πn

Show solutions in decimal form

x = 0.29725 \dots + 2 πn, x = π - 0.29725 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)+1=4sin(x) ?
The general solution for 2sin^2(x)+1=4sin(x) is x=0.29725…+2pin,x=pi-0.29725…+2pin