What is the general solution for cos(x)+9sin^2(x)=1 ?

The general solution for cos(x)+9sin^2(x)=1 is x=2.66571…+2pin,x=-2.66571…+2pin,x=2pin

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cos(x)+9sin^2(x)=1

Solution

cos (x) + 9 sin^{2} (x) = 1

Solution

x = 2.66571 \dots + 2 πn, x = - 2.66571 \dots + 2 πn, x = 2 πn

Degrees

x = 152.73395 \dots^{\circ} + 36 0^{\circ} n, x = - 152.73395 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (x) + 9 sin^{2} (x) = 1

Subtract

1

from both sides

cos (x) + 9 sin^{2} (x) - 1 = 0

Rewrite using trig identities

- 1 + cos (x) + 9 sin^{2} (x)

Use the Pythagorean identity:

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

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

= - 1 + cos (x) + 9 (1 - cos^{2} (x))

Simplify

- 1 + cos (x) + 9 (1 - cos^{2} (x)) : cos (x) - 9 cos^{2} (x) + 8

- 1 + cos (x) + 9 (1 - cos^{2} (x))

Expand

9 (1 - cos^{2} (x)) : 9 - 9 cos^{2} (x)

9 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (x)

= 9 \cdot 1 - 9 cos^{2} (x)

Multiply the numbers:

9 \cdot 1 = 9

= 9 - 9 cos^{2} (x)

= - 1 + cos (x) + 9 - 9 cos^{2} (x)

Simplify

- 1 + cos (x) + 9 - 9 cos^{2} (x) : cos (x) - 9 cos^{2} (x) + 8

- 1 + cos (x) + 9 - 9 cos^{2} (x)

Group like terms

= cos (x) - 9 cos^{2} (x) - 1 + 9

Add/Subtract the numbers:

- 1 + 9 = 8

= cos (x) - 9 cos^{2} (x) + 8

= cos (x) - 9 cos^{2} (x) + 8

= cos (x) - 9 cos^{2} (x) + 8

8 + cos (x) - 9 cos^{2} (x) = 0

Solve by substitution

8 + cos (x) - 9 cos^{2} (x) = 0

Let:

cos (x) = u

8 + u - 9 u^{2} = 0

8 + u - 9 u^{2} = 0 : u = - \frac{8}{9}, u = 1

8 + u - 9 u^{2} = 0

Write in the standard form

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

- 9 u^{2} + u + 8 = 0

Solve with the quadratic formula

- 9 u^{2} + u + 8 = 0

Quadratic Equation Formula:

For

a = - 9, b = 1, c = 8

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 9 ) \cdot 8}{2 ( - 9 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 9 ) \cdot 8}{2 ( - 9 )}

1^{2} - 4 (- 9) \cdot 8 = 17

1^{2} - 4 (- 9) \cdot 8

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 9) \cdot 8

Apply rule

- (- a) = a

= 1 + 4 \cdot 9 \cdot 8

Multiply the numbers:

4 \cdot 9 \cdot 8 = 288

= 1 + 288

Add the numbers:

1 + 288 = 289

= 289

Factor the number:

289 = 1 7^{2}

= 1 7^{2}

Apply radical rule:

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- 1 \pm 17}{2 ( - 9 )}

Separate the solutions

u_{1} = \frac{- 1 + 17}{2 ( - 9 )}, u_{2} = \frac{- 1 - 17}{2 ( - 9 )}

u = \frac{- 1 + 17}{2 ( - 9 )} : - \frac{8}{9}

\frac{- 1 + 17}{2 ( - 9 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 17}{- 2 \cdot 9}

Add/Subtract the numbers:

- 1 + 17 = 16

= \frac{16}{- 2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{16}{- 18}

Apply the fraction rule:

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

= - \frac{16}{18}

Cancel the common factor:

2

= - \frac{8}{9}

u = \frac{- 1 - 17}{2 ( - 9 )} : 1

\frac{- 1 - 17}{2 ( - 9 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 17}{- 2 \cdot 9}

Subtract the numbers:

- 1 - 17 = - 18

= \frac{- 18}{- 2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{- 18}{- 18}

Apply the fraction rule:

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

= \frac{18}{18}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - \frac{8}{9}, u = 1

Substitute back

u = cos (x)

cos (x) = - \frac{8}{9}, cos (x) = 1

cos (x) = - \frac{8}{9}, cos (x) = 1

cos (x) = - \frac{8}{9} : x = arccos (- \frac{8}{9}) + 2 πn, x = - arccos (- \frac{8}{9}) + 2 πn

cos (x) = - \frac{8}{9}

Apply trig inverse properties

cos (x) = - \frac{8}{9}

General solutions for

cos (x) = - \frac{8}{9}

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

x = arccos (- \frac{8}{9}) + 2 πn, x = - arccos (- \frac{8}{9}) + 2 πn

x = arccos (- \frac{8}{9}) + 2 πn, x = - arccos (- \frac{8}{9}) + 2 πn

cos (x) = 1 : x = 2 πn

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Combine all the solutions

x = arccos (- \frac{8}{9}) + 2 πn, x = - arccos (- \frac{8}{9}) + 2 πn, x = 2 πn

Show solutions in decimal form

x = 2.66571 \dots + 2 πn, x = - 2.66571 \dots + 2 πn, x = 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(x)+9sin^2(x)=1 ?
The general solution for cos(x)+9sin^2(x)=1 is x=2.66571…+2pin,x=-2.66571…+2pin,x=2pin