What is the general solution for 4sin^2(x)=5-2cos(x) ?

The general solution for 4sin^2(x)=5-2cos(x) is No Solution for x\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4sin^2(x)=5-2cos(x)

Solution

4 sin^{2} (x) = 5 - 2 cos (x)

Solution

No Solution for x \in R

Solution steps

4 sin^{2} (x) = 5 - 2 cos (x)

Subtract

5 - 2 cos (x)

from both sides

4 sin^{2} (x) - 5 + 2 cos (x) = 0

Rewrite using trig identities

- 5 + 2 cos (x) + 4 sin^{2} (x)

Use the Pythagorean identity:

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

sin^{2} (x) = 1 - cos^{2} (x)

= - 5 + 2 cos (x) + 4 (1 - cos^{2} (x))

Simplify

- 5 + 2 cos (x) + 4 (1 - cos^{2} (x)) : 2 cos (x) - 4 cos^{2} (x) - 1

- 5 + 2 cos (x) + 4 (1 - cos^{2} (x))

Expand

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

4 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

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

Multiply the numbers:

4 \cdot 1 = 4

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

= - 5 + 2 cos (x) + 4 - 4 cos^{2} (x)

Simplify

- 5 + 2 cos (x) + 4 - 4 cos^{2} (x) : 2 cos (x) - 4 cos^{2} (x) - 1

- 5 + 2 cos (x) + 4 - 4 cos^{2} (x)

Group like terms

= 2 cos (x) - 4 cos^{2} (x) - 5 + 4

Add/Subtract the numbers:

- 5 + 4 = - 1

= 2 cos (x) - 4 cos^{2} (x) - 1

= 2 cos (x) - 4 cos^{2} (x) - 1

= 2 cos (x) - 4 cos^{2} (x) - 1

- 1 + 2 cos (x) - 4 cos^{2} (x) = 0

Solve by substitution

- 1 + 2 cos (x) - 4 cos^{2} (x) = 0

Let:

cos (x) = u

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

- 1 + 2 u - 4 u^{2} = 0 : u = \frac{1}{4} - i \frac{3}{4}, u = \frac{1}{4} + i \frac{3}{4}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

Simplify

2^{2} - 4 (- 4) (- 1) : 23 i

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 2^{2} - 16

Apply imaginary number rule:

- a = i a

= i 16 - 2^{2}

- 2^{2} + 16 = 23

- 2^{2} + 16

2^{2} = 4

= - 4 + 16

Add/Subtract the numbers:

- 4 + 16 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

= 23 i

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

Separate the solutions

u_{1} = \frac{- 2 + 2 3 i}{2 ( - 4 )}, u_{2} = \frac{- 2 - 2 3 i}{2 ( - 4 )}

u = \frac{- 2 + 2 3 i}{2 ( - 4 )} : \frac{1}{4} - i \frac{3}{4}

\frac{- 2 + 2 3 i}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 + 2 3 i}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2 + 2 3 i}{- 8}

Apply the fraction rule:

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

= - \frac{- 2 + 2 3 i}{8}

Cancel

\frac{- 2 + 2 3 i}{8} : \frac{- 1 + 3 i}{4}

\frac{- 2 + 2 3 i}{8}

Factor

- 2 + 23 i : 2 (- 1 + 3 i)

- 2 + 23 i

Rewrite as

= - 2 \cdot 1 + 23 i

Factor out common term

2

= 2 (- 1 + 3 i)

= \frac{2 ( - 1 + 3 i )}{8}

Cancel the common factor:

2

= \frac{- 1 + 3 i}{4}

= - \frac{- 1 + 3 i}{4}

Rewrite

- \frac{- 1 + 3 i}{4}

in standard complex form:

\frac{1}{4} - \frac{3}{4} i

- \frac{- 1 + 3 i}{4}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + 3 i}{4} = - (- \frac{1}{4}) - (\frac{3 i}{4})

= - (- \frac{1}{4}) - (\frac{3 i}{4})

Remove parentheses:

(a) = a, - (- a) = a

= \frac{1}{4} - \frac{3 i}{4}

= \frac{1}{4} - \frac{3}{4} i

u = \frac{- 2 - 2 3 i}{2 ( - 4 )} : \frac{1}{4} + i \frac{3}{4}

\frac{- 2 - 2 3 i}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 - 2 3 i}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2 - 2 3 i}{- 8}

Apply the fraction rule:

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

= - \frac{- 2 - 2 3 i}{8}

Cancel

\frac{- 2 - 2 3 i}{8} : - \frac{1 + 3 i}{4}

\frac{- 2 - 2 3 i}{8}

Factor

- 2 - 23 i : - 2 (1 + 3 i)

- 2 - 23 i

Rewrite as

= - 2 \cdot 1 - 23 i

Factor out common term

2

= - 2 (1 + 3 i)

= - \frac{2 ( 1 + 3 i )}{8}

Cancel the common factor:

2

= - \frac{1 + 3 i}{4}

= - (- \frac{1 + 3 i}{4})

Apply rule

- (- a) = a

= \frac{1 + 3 i}{4}

Rewrite

\frac{1 + 3 i}{4}

in standard complex form:

\frac{1}{4} + \frac{3}{4} i

\frac{1 + 3 i}{4}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{1 + 3 i}{4} = \frac{1}{4} + \frac{3 i}{4}

= \frac{1}{4} + \frac{3 i}{4}

= \frac{1}{4} + \frac{3}{4} i

The solutions to the quadratic equation are:

u = \frac{1}{4} - i \frac{3}{4}, u = \frac{1}{4} + i \frac{3}{4}

Substitute back

u = cos (x)

cos (x) = \frac{1}{4} - i \frac{3}{4}, cos (x) = \frac{1}{4} + i \frac{3}{4}

cos (x) = \frac{1}{4} - i \frac{3}{4}, cos (x) = \frac{1}{4} + i \frac{3}{4}

cos (x) = \frac{1}{4} - i \frac{3}{4} :

No Solution

cos (x) = \frac{1}{4} - i \frac{3}{4}

No Solution

cos (x) = \frac{1}{4} + i \frac{3}{4} :

No Solution

cos (x) = \frac{1}{4} + i \frac{3}{4}

No Solution

Combine all the solutions

No Solution for x \in R

Popular Examples

sin(x+pi/6)=-1

sin (x + \frac{π}{6}) = - 1

sec(x)= 4/3

sec (x) = \frac{4}{3}

4cos(x)+3cos(x)=5

4 cos (x) + 3 cos (x) = 5

4sin(2x)tan(x)-tan(x)=0

4 sin (2 x) tan (x) - tan (x) = 0

3sin(x)=sqrt(3)*sin(x)+3cos(x)+3sin(x)

3 sin (x) = 3 \cdot sin (x) + 3 cos (x) + 3 sin (x)

Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)=5-2cos(x) ?
The general solution for 4sin^2(x)=5-2cos(x) is No Solution for x\in\mathbb{R}