What is the general solution for 4sin^2(x)+cos(x)+1=0 ?

The general solution for 4sin^2(x)+cos(x)+1=0 is x=pi+2pin

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Popular >

4sin^2(x)+cos(x)+1=0

Solution

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

Solution

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

1 + 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)

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

Simplify

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

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

Group like terms

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

Add the numbers:

1 + 4 = 5

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

5 + u - 4 u^{2} = 0 : u = - 1, u = \frac{5}{4}

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

Write in the standard form

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

- 4 u^{2} + u + 5 = 0

Solve with the quadratic formula

- 4 u^{2} + u + 5 = 0

Quadratic Equation Formula:

For

a = - 4, b = 1, c = 5

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

1^{2} - 4 (- 4) \cdot 5 = 9

1^{2} - 4 (- 4) \cdot 5

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 4) \cdot 5

Apply rule

- (- a) = a

= 1 + 4 \cdot 4 \cdot 5

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 1 + 80

Add the numbers:

1 + 80 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- 1 \pm 9}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 1 + 9}{2 ( - 4 )}, u_{2} = \frac{- 1 - 9}{2 ( - 4 )}

u = \frac{- 1 + 9}{2 ( - 4 )} : - 1

\frac{- 1 + 9}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 9}{- 2 \cdot 4}

Add/Subtract the numbers:

- 1 + 9 = 8

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{- 8}

Apply the fraction rule:

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

= - \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- 1 - 9}{2 ( - 4 )} : \frac{5}{4}

\frac{- 1 - 9}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 9}{- 2 \cdot 4}

Subtract the numbers:

- 1 - 9 = - 10

= \frac{- 10}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 10}{- 8}

Apply the fraction rule:

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

= \frac{10}{8}

Cancel the common factor:

2

= \frac{5}{4}

The solutions to the quadratic equation are:

u = - 1, u = \frac{5}{4}

Substitute back

u = cos (x)

cos (x) = - 1, cos (x) = \frac{5}{4}

cos (x) = - 1, cos (x) = \frac{5}{4}

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

cos (x) = \frac{5}{4} :

No Solution

cos (x) = \frac{5}{4}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)+cos(x)+1=0 ?
The general solution for 4sin^2(x)+cos(x)+1=0 is x=pi+2pin