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

The general solution for 4sin^2(x)+4sin(x)=1 is x=0.20861…+2pin,x=pi-0.20861…+2pin

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Popular >

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

Solution

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

Solution

x = 0.20861 \dots + 2 πn, x = π - 0.20861 \dots + 2 πn

Degrees

x = 11.95285 \dots^{\circ} + 36 0^{\circ} n, x = 168.04714 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

4 u^{2} + 4 u = 1

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

4 u^{2} + 4 u = 1

Move

1

to the left side

4 u^{2} + 4 u = 1

Subtract

1

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

4^{2} - 4 \cdot 4 (- 1) = 42

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} + 16

4^{2} = 16

= 16 + 16

Add the numbers:

16 + 16 = 32

= 32

Prime factorization of

32 : 2^{5}

32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

Apply exponent rule:

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

= 2^{4} \cdot 2

Apply radical rule:

= 2 2^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

Refine

= 42

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

Separate the solutions

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

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

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

Multiply the numbers:

2 \cdot 4 = 8

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

Factor

- 4 + 42 : 4 (- 1 + 2)

- 4 + 42

Rewrite as

= - 4 \cdot 1 + 42

Factor out common term

4

= 4 (- 1 + 2)

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

Cancel the common factor:

4

= \frac{- 1 + 2}{2}

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

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

Multiply the numbers:

2 \cdot 4 = 8

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

Factor

- 4 - 42 : - 4 (1 + 2)

- 4 - 42

Rewrite as

= - 4 \cdot 1 - 42

Factor out common term

4

= - 4 (1 + 2)

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

Cancel the common factor:

4

= - \frac{1 + 2}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

x = arcsin (\frac{- 1 + 2}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 2}{2}) + 2 πn

x = arcsin (\frac{- 1 + 2}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 2}{2}) + 2 πn

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (\frac{- 1 + 2}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 2}{2}) + 2 πn

Show solutions in decimal form

x = 0.20861 \dots + 2 πn, x = π - 0.20861 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)+4sin(x)=1 ?
The general solution for 4sin^2(x)+4sin(x)=1 is x=0.20861…+2pin,x=pi-0.20861…+2pin