What is the general solution for 3cos^2(x)+sin(x)+1=0 ?

The general solution for 3cos^2(x)+sin(x)+1=0 is x=(3pi)/2+2pin

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Popular >

3cos^2(x)+sin(x)+1=0

Solution

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

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

1 + sin (x) + 3 cos^{2} (x)

Use the Pythagorean identity:

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

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

= 1 + sin (x) + 3 (1 - sin^{2} (x))

Simplify

1 + sin (x) + 3 (1 - sin^{2} (x)) : sin (x) - 3 sin^{2} (x) + 4

1 + sin (x) + 3 (1 - sin^{2} (x))

Expand

3 (1 - sin^{2} (x)) : 3 - 3 sin^{2} (x)

3 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (x)

= 3 \cdot 1 - 3 sin^{2} (x)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 sin^{2} (x)

= 1 + sin (x) + 3 - 3 sin^{2} (x)

Simplify

1 + sin (x) + 3 - 3 sin^{2} (x) : sin (x) - 3 sin^{2} (x) + 4

1 + sin (x) + 3 - 3 sin^{2} (x)

Group like terms

= sin (x) - 3 sin^{2} (x) + 1 + 3

Add the numbers:

1 + 3 = 4

= sin (x) - 3 sin^{2} (x) + 4

= sin (x) - 3 sin^{2} (x) + 4

= sin (x) - 3 sin^{2} (x) + 4

4 + sin (x) - 3 sin^{2} (x) = 0

Solve by substitution

4 + sin (x) - 3 sin^{2} (x) = 0

Let:

sin (x) = u

4 + u - 3 u^{2} = 0

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

4 + u - 3 u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 3, b = 1, c = 4

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

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

1^{2} - 4 (- 3) \cdot 4 = 7

1^{2} - 4 (- 3) \cdot 4

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 3) \cdot 4

Apply rule

- (- a) = a

= 1 + 4 \cdot 3 \cdot 4

Multiply the numbers:

4 \cdot 3 \cdot 4 = 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 1 \pm 7}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- 1 + 7}{2 ( - 3 )}, u_{2} = \frac{- 1 - 7}{2 ( - 3 )}

u = \frac{- 1 + 7}{2 ( - 3 )} : - 1

\frac{- 1 + 7}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 7}{- 2 \cdot 3}

Add/Subtract the numbers:

- 1 + 7 = 6

= \frac{6}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{- 6}

Apply the fraction rule:

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

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- 1 - 7}{2 ( - 3 )} : \frac{4}{3}

\frac{- 1 - 7}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 7}{- 2 \cdot 3}

Subtract the numbers:

- 1 - 7 = - 8

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 8}{- 6}

Apply the fraction rule:

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

= \frac{8}{6}

Cancel the common factor:

2

= \frac{4}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

sin (x) = - 1, sin (x) = \frac{4}{3}

sin (x) = - 1, sin (x) = \frac{4}{3}

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

General solutions for

sin (x) = - 1

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

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = \frac{4}{3} :

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

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sin(x)=7

sin (x) = 7

sin^2(x)cos^2(x)=(2-sqrt(2))/(16)

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)+sin(x)+1=0 ?
The general solution for 3cos^2(x)+sin(x)+1=0 is x=(3pi)/2+2pin