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

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

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

Solution

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

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 1, b = 2, c = 3

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

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

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

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

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

Add the numbers:

4 + 12 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 2 + 4 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{- 2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

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

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 2 - 4 = - 6

= \frac{- 6}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 6}{- 2}

Apply the fraction rule:

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

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

The solutions to the quadratic equation are:

u = - 1, u = 3

Substitute back

u = sin (x)

sin (x) = - 1, sin (x) = 3

sin (x) = - 1, sin (x) = 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) = 3 :

No Solution

sin (x) = 3

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Popular Examples

2sin^2(2x+pi)-sin(2x+pi)-1=0

2 sin^{2} (2 x + π) - sin (2 x + π) - 1 = 0

cos(x)=9

cos (x) = 9

sec(θ)= 3/5

sec (θ) = \frac{3}{5}

cos(3x)+2cos(x)=0

cos (3 x) + 2 cos (x) = 0

sin(x+pi/2)-cos(x+(3pi)/2)=0

sin (x + \frac{π}{2}) - cos (x + \frac{3 π}{2}) = 0

Frequently Asked Questions (FAQ)

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