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

The general solution for 2cos(x)+sin^2(x)=0 is x=1.99787…+2pin,x=-1.99787…+2pin

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Popular >

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

Solution

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

Solution

x = 1.99787 \dots + 2 πn, x = - 1.99787 \dots + 2 πn

Degrees

x = 114.46980 \dots^{\circ} + 36 0^{\circ} n, x = - 114.46980 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

- u^{2} + 2 u + 1 = 0

Solve with the quadratic formula

- u^{2} + 2 u + 1 = 0

Quadratic Equation Formula:

For

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

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

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

2^{2} - 4 (- 1) \cdot 1 = 22

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

Apply radical rule:

n ab = n a n b

= 2 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 22

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

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

Cancel

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

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

Factor

- 2 + 22 : 2 (- 1 + 2)

- 2 + 22

Rewrite as

= - 2 \cdot 1 + 22

Factor out common term

2

= 2 (- 1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= - 1 + 2

= - (2 - 1)

Distribute parentheses

= - (- 1) - (2)

Apply minus-plus rules

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

= 1 - 2

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

- 2 - 22 = - (2 + 22)

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

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2

The solutions to the quadratic equation are:

u = 1 - 2, u = 1 + 2

Substitute back

u = cos (x)

cos (x) = 1 - 2, cos (x) = 1 + 2

cos (x) = 1 - 2, cos (x) = 1 + 2

cos (x) = 1 - 2 : x = arccos (1 - 2) + 2 πn, x = - arccos (1 - 2) + 2 πn

cos (x) = 1 - 2

Apply trig inverse properties

cos (x) = 1 - 2

General solutions for

cos (x) = 1 - 2

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (1 - 2) + 2 πn, x = - arccos (1 - 2) + 2 πn

x = arccos (1 - 2) + 2 πn, x = - arccos (1 - 2) + 2 πn

cos (x) = 1 + 2 :

No Solution

cos (x) = 1 + 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (1 - 2) + 2 πn, x = - arccos (1 - 2) + 2 πn

Show solutions in decimal form

x = 1.99787 \dots + 2 πn, x = - 1.99787 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(x)+sin^2(x)=0 ?
The general solution for 2cos(x)+sin^2(x)=0 is x=1.99787…+2pin,x=-1.99787…+2pin