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

The general solution for cos(x)+2sin(x)+3=0 is No Solution for x\in\mathbb{R}

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

2 sin (x)

from both sides

cos (x) + 3 = - 2 sin (x)

Square both sides

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

Subtract

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

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= (3 + cos (x))^{2} - 4 (1 - cos^{2} (x))

Simplify

(3 + cos (x))^{2} - 4 (1 - cos^{2} (x)) : 5 cos^{2} (x) + 6 cos (x) + 5

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

(3 + cos (x))^{2} : 9 + 6 cos (x) + cos^{2} (x)

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 3, b = cos (x)

= 3^{2} + 2 \cdot 3 cos (x) + cos^{2} (x)

Simplify

3^{2} + 2 \cdot 3 cos (x) + cos^{2} (x) : 9 + 6 cos (x) + cos^{2} (x)

3^{2} + 2 \cdot 3 cos (x) + cos^{2} (x)

3^{2} = 9

= 9 + 2 \cdot 3 cos (x) + cos^{2} (x)

Multiply the numbers:

2 \cdot 3 = 6

= 9 + 6 cos (x) + cos^{2} (x)

= 9 + 6 cos (x) + cos^{2} (x)

= 9 + 6 cos (x) + cos^{2} (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)

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 = 4

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

= 9 + 6 cos (x) + cos^{2} (x) - 4 + 4 cos^{2} (x)

Simplify

9 + 6 cos (x) + cos^{2} (x) - 4 + 4 cos^{2} (x) : 5 cos^{2} (x) + 6 cos (x) + 5

9 + 6 cos (x) + cos^{2} (x) - 4 + 4 cos^{2} (x)

Group like terms

= 6 cos (x) + cos^{2} (x) + 4 cos^{2} (x) + 9 - 4

Add similar elements:

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

= 6 cos (x) + 5 cos^{2} (x) + 9 - 4

Add/Subtract the numbers:

9 - 4 = 5

= 5 cos^{2} (x) + 6 cos (x) + 5

= 5 cos^{2} (x) + 6 cos (x) + 5

= 5 cos^{2} (x) + 6 cos (x) + 5

5 + 5 cos^{2} (x) + 6 cos (x) = 0

Solve by substitution

5 + 5 cos^{2} (x) + 6 cos (x) = 0

Let:

cos (x) = u

5 + 5 u^{2} + 6 u = 0

5 + 5 u^{2} + 6 u = 0 : u = - \frac{3}{5} + i \frac{4}{5}, u = - \frac{3}{5} - i \frac{4}{5}

5 + 5 u^{2} + 6 u = 0

Write in the standard form

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

5 u^{2} + 6 u + 5 = 0

Solve with the quadratic formula

5 u^{2} + 6 u + 5 = 0

Quadratic Equation Formula:

For

a = 5, b = 6, c = 5

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 5 \cdot 5}{2 \cdot 5}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 5 \cdot 5}{2 \cdot 5}

Simplify

6^{2} - 4 \cdot 5 \cdot 5 : 8 i

6^{2} - 4 \cdot 5 \cdot 5

Multiply the numbers:

4 \cdot 5 \cdot 5 = 100

= 6^{2} - 100

Apply imaginary number rule:

- a = i a

= i 100 - 6^{2}

- 6^{2} + 100 = 8

- 6^{2} + 100

6^{2} = 36

= - 36 + 100

Add/Subtract the numbers:

- 36 + 100 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

= 8 i

u_{1, 2} = \frac{- 6 \pm 8 i}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- 6 + 8 i}{2 \cdot 5}, u_{2} = \frac{- 6 - 8 i}{2 \cdot 5}

u = \frac{- 6 + 8 i}{2 \cdot 5} : - \frac{3}{5} + i \frac{4}{5}

\frac{- 6 + 8 i}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 6 + 8 i}{10}

Factor

- 6 + 8 i : 2 (- 3 + 4 i)

- 6 + 8 i

Rewrite as

= - 2 \cdot 3 + 2 \cdot 4 i

Factor out common term

2

= 2 (- 3 + 4 i)

= \frac{2 ( - 3 + 4 i )}{10}

Cancel the common factor:

2

= \frac{- 3 + 4 i}{5}

Rewrite

\frac{- 3 + 4 i}{5}

in standard complex form:

- \frac{3}{5} + \frac{4}{5} i

\frac{- 3 + 4 i}{5}

Apply the fraction rule:

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

\frac{- 3 + 4 i}{5} = - \frac{3}{5} + \frac{4 i}{5}

= - \frac{3}{5} + \frac{4 i}{5}

= - \frac{3}{5} + \frac{4}{5} i

u = \frac{- 6 - 8 i}{2 \cdot 5} : - \frac{3}{5} - i \frac{4}{5}

\frac{- 6 - 8 i}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 6 - 8 i}{10}

Factor

- 6 - 8 i : - 2 (3 + 4 i)

- 6 - 8 i

Rewrite as

= - 2 \cdot 3 - 2 \cdot 4 i

Factor out common term

2

= - 2 (3 + 4 i)

= - \frac{2 ( 3 + 4 i )}{10}

Cancel the common factor:

2

= - \frac{3 + 4 i}{5}

Rewrite

- \frac{3 + 4 i}{5}

in standard complex form:

- \frac{3}{5} - \frac{4}{5} i

- \frac{3 + 4 i}{5}

Apply the fraction rule:

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

\frac{3 + 4 i}{5} = - (\frac{3}{5}) - (\frac{4 i}{5})

= - (\frac{3}{5}) - (\frac{4 i}{5})

Remove parentheses:

(a) = a

= - \frac{3}{5} - \frac{4 i}{5}

= - \frac{3}{5} - \frac{4}{5} i

The solutions to the quadratic equation are:

u = - \frac{3}{5} + i \frac{4}{5}, u = - \frac{3}{5} - i \frac{4}{5}

Substitute back

u = cos (x)

cos (x) = - \frac{3}{5} + i \frac{4}{5}, cos (x) = - \frac{3}{5} - i \frac{4}{5}

cos (x) = - \frac{3}{5} + i \frac{4}{5}, cos (x) = - \frac{3}{5} - i \frac{4}{5}

cos (x) = - \frac{3}{5} + i \frac{4}{5} :

No Solution

cos (x) = - \frac{3}{5} + i \frac{4}{5}

No Solution

cos (x) = - \frac{3}{5} - i \frac{4}{5} :

No Solution

cos (x) = - \frac{3}{5} - i \frac{4}{5}

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

Remove the ones that don't agree with the equation.

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for cos(x)+2sin(x)+3=0 ?
The general solution for cos(x)+2sin(x)+3=0 is No Solution for x\in\mathbb{R}