What is the general solution for 2cos(2x)+sin(x)-4=0 ?

The general solution for 2cos(2x)+sin(x)-4=0 is No Solution for x\in\mathbb{R}

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Popular Trigonometry >

2cos(2x)+sin(x)-4=0

Solution

2 cos (2 x) + sin (x) - 4 = 0

Solution

No Solution for x \in R

Solution steps

2 cos (2 x) + sin (x) - 4 = 0

Rewrite using trig identities

- 4 + sin (x) + 2 cos (2 x)

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Simplify

2 \cdot 1 - 2 \cdot 2 sin^{2} (x) : 2 - 4 sin^{2} (x)

2 \cdot 1 - 2 \cdot 2 sin^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 \cdot 2 sin^{2} (x)

Multiply the numbers:

2 \cdot 2 = 4

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

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

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

Simplify

- 4 + sin (x) + 2 - 4 sin^{2} (x) : sin (x) - 4 sin^{2} (x) - 2

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

Group like terms

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

Add/Subtract the numbers:

- 4 + 2 = - 2

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

- 2 + u - 4 u^{2} = 0 : u = \frac{1}{8} - i \frac{31}{8}, u = \frac{1}{8} + i \frac{31}{8}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

Simplify

1^{2} - 4 (- 4) (- 2) : 31 i

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 4) (- 2)

Apply rule

- (- a) = a

= 1 - 4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 1 - 32

Subtract the numbers:

1 - 32 = - 31

= - 31

Apply radical rule:

- a = - 1 a

- 31 = - 1 31

= - 1 31

Apply imaginary number rule:

- 1 = i

= 31 i

u_{1, 2} = \frac{- 1 \pm 31 i}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 1 + 31 i}{2 ( - 4 )}, u_{2} = \frac{- 1 - 31 i}{2 ( - 4 )}

u = \frac{- 1 + 31 i}{2 ( - 4 )} : \frac{1}{8} - i \frac{31}{8}

\frac{- 1 + 31 i}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 31 i}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 1 + 31 i}{- 8}

Apply the fraction rule:

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

= - \frac{- 1 + 31 i}{8}

Rewrite

- \frac{- 1 + 31 i}{8}

in standard complex form:

\frac{1}{8} - \frac{31}{8} i

- \frac{- 1 + 31 i}{8}

Apply the fraction rule:

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

\frac{- 1 + 31 i}{8} = - (- \frac{1}{8}) - (\frac{31 i}{8})

= - (- \frac{1}{8}) - (\frac{31 i}{8})

Remove parentheses:

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

= \frac{1}{8} - \frac{31 i}{8}

= \frac{1}{8} - \frac{31}{8} i

u = \frac{- 1 - 31 i}{2 ( - 4 )} : \frac{1}{8} + i \frac{31}{8}

\frac{- 1 - 31 i}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 31 i}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 1 - 31 i}{- 8}

Apply the fraction rule:

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

= - \frac{- 1 - 31 i}{8}

Rewrite

- \frac{- 1 - 31 i}{8}

in standard complex form:

\frac{1}{8} + \frac{31}{8} i

- \frac{- 1 - 31 i}{8}

Apply the fraction rule:

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

\frac{- 1 - 31 i}{8} = - (- \frac{1}{8}) - (- \frac{31 i}{8})

= - (- \frac{1}{8}) - (- \frac{31 i}{8})

Apply rule

- (- a) = a

= \frac{1}{8} + \frac{31 i}{8}

= \frac{1}{8} + \frac{31}{8} i

The solutions to the quadratic equation are:

u = \frac{1}{8} - i \frac{31}{8}, u = \frac{1}{8} + i \frac{31}{8}

Substitute back

u = sin (x)

sin (x) = \frac{1}{8} - i \frac{31}{8}, sin (x) = \frac{1}{8} + i \frac{31}{8}

sin (x) = \frac{1}{8} - i \frac{31}{8}, sin (x) = \frac{1}{8} + i \frac{31}{8}

sin (x) = \frac{1}{8} - i \frac{31}{8} :

No Solution

sin (x) = \frac{1}{8} - i \frac{31}{8}

No Solution

sin (x) = \frac{1}{8} + i \frac{31}{8} :

No Solution

sin (x) = \frac{1}{8} + i \frac{31}{8}

No Solution

Combine all the solutions

No Solution for x \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2x)+sin(x)-4=0 ?
The general solution for 2cos(2x)+sin(x)-4=0 is No Solution for x\in\mathbb{R}