What is the general solution for cos(x)=-sin^2(x)-1,0<= x<= 2pi ?

Q: What is the general solution for cos(x)=-sin^2(x)-1,0<= x<= 2pi ?

The general solution for cos(x)=-sin^2(x)-1,0<= x<= 2pi is x=pi

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cos(x)=-sin^2(x)-1,0<= x<= 2pi

Solution

cos (x) = - sin^{2} (x) - 1, 0 \leq x \leq 2 π

Solution

x = π

Degrees

x = 18 0^{\circ}

Solution steps

cos (x) = - sin^{2} (x) - 1, 0 \leq x \leq 2 π

Subtract

- sin^{2} (x) - 1

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Group like terms

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

Add the numbers:

1 + 1 = 2

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

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

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

Solve by substitution

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

Let:

cos (x) = u

2 + u - u^{2} = 0

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

2 + u - u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 2

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 1 + 3 = 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{- 1 - 3}{2 ( - 1 )} : 2

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 1 - 3 = - 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{- 2}

Apply the fraction rule:

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

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

The solutions to the quadratic equation are:

u = - 1, u = 2

Substitute back

u = cos (x)

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

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

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

cos (x) = - 1, 0 \leq x \leq 2 π

General solutions for

cos (x) = - 1

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

x = π + 2 πn

x = π + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = π

cos (x) = 2, 0 \leq x \leq 2 π :

No Solution

cos (x) = 2, 0 \leq x \leq 2 π

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = π

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Frequently Asked Questions (FAQ)

What is the general solution for cos(x)=-sin^2(x)-1,0<= x<= 2pi ?
The general solution for cos(x)=-sin^2(x)-1,0<= x<= 2pi is x=pi