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

The general solution for cos^2(θ)+2sin(θ)+1=0 is θ=-0.82132…+2pin,θ=pi+0.82132…+2pin

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Popular >

cos^2(θ)+2sin(θ)+1=0

Solution

cos^{2} (θ) + 2 sin (θ) + 1 = 0

Solution

θ = - 0.82132 \dots + 2 πn, θ = π + 0.82132 \dots + 2 πn

Degrees

θ = - 47.05859 \dots^{\circ} + 36 0^{\circ} n, θ = 227.05859 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (θ) + 2 sin (θ) + 1 = 0

Rewrite using trig identities

1 + cos^{2} (θ) + 2 sin (θ)

Use the Pythagorean identity:

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

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

= 1 + 1 - sin^{2} (θ) + 2 sin (θ)

Simplify

= 2 sin (θ) - sin^{2} (θ) + 2

2 - sin^{2} (θ) + 2 sin (θ) = 0

Solve by substitution

2 - sin^{2} (θ) + 2 sin (θ) = 0

Let:

sin (θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

2^{2} - 4 (- 1) \cdot 2 = 23

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

Add the numbers:

4 + 8 = 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

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

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

Cancel

\frac{- 2 + 2 3}{2} : 3 - 1

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

Factor

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

- 2 + 23

Rewrite as

= - 2 \cdot 1 + 23

Factor out common term

2

= 2 (- 1 + 3)

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

Divide the numbers:

\frac{2}{2} = 1

= - 1 + 3

= - (3 - 1)

Distribute parentheses

= - (- 1) - (3)

Apply minus-plus rules

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

= 1 - 3

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

- 2 - 23 = - (2 + 23)

= \frac{2 + 2 3}{2}

Factor

2 + 23 : 2 (1 + 3)

2 + 23

Rewrite as

= 2 \cdot 1 + 23

Factor out common term

2

= 2 (1 + 3)

= \frac{2 ( 1 + 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 1 + 3

The solutions to the quadratic equation are:

u = 1 - 3, u = 1 + 3

Substitute back

u = sin (θ)

sin (θ) = 1 - 3, sin (θ) = 1 + 3

sin (θ) = 1 - 3, sin (θ) = 1 + 3

sin (θ) = 1 - 3 : θ = arcsin (1 - 3) + 2 πn, θ = π + arcsin (- 1 + 3) + 2 πn

sin (θ) = 1 - 3

Apply trig inverse properties

sin (θ) = 1 - 3

General solutions for

sin (θ) = 1 - 3

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (1 - 3) + 2 πn, θ = π + arcsin (- 1 + 3) + 2 πn

θ = arcsin (1 - 3) + 2 πn, θ = π + arcsin (- 1 + 3) + 2 πn

sin (θ) = 1 + 3 :

No Solution

sin (θ) = 1 + 3

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = arcsin (1 - 3) + 2 πn, θ = π + arcsin (- 1 + 3) + 2 πn

Show solutions in decimal form

θ = - 0.82132 \dots + 2 πn, θ = π + 0.82132 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(θ)+2sin(θ)+1=0 ?
The general solution for cos^2(θ)+2sin(θ)+1=0 is θ=-0.82132…+2pin,θ=pi+0.82132…+2pin