What is the general solution for -9sin^2(x)-5cos(x)+5=0 ?

The general solution for -9sin^2(x)-5cos(x)+5=0 is x=2pin,x=2.03135…+2pin,x=-2.03135…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-9sin^2(x)-5cos(x)+5=0

Solution

- 9 sin^{2} (x) - 5 cos (x) + 5 = 0

Solution

x = 2 πn, x = 2.03135 \dots + 2 πn, x = - 2.03135 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 116.38779 \dots^{\circ} + 36 0^{\circ} n, x = - 116.38779 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 9 sin^{2} (x) - 5 cos (x) + 5 = 0

Rewrite using trig identities

5 - 5 cos (x) - 9 sin^{2} (x)

Use the Pythagorean identity:

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

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

= 5 - 5 cos (x) - 9 (1 - cos^{2} (x))

Simplify

5 - 5 cos (x) - 9 (1 - cos^{2} (x)) : 9 cos^{2} (x) - 5 cos (x) - 4

5 - 5 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)

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

9 \cdot 1 = 9

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

5 - 9 = - 4

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

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

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

- 4 - 5 cos (x) + 9 cos^{2} (x) = 0

Solve by substitution

- 4 - 5 cos (x) + 9 cos^{2} (x) = 0

Let:

cos (x) = u

- 4 - 5 u + 9 u^{2} = 0

- 4 - 5 u + 9 u^{2} = 0 : u = 1, u = - \frac{4}{9}

- 4 - 5 u + 9 u^{2} = 0

Write in the standard form

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

9 u^{2} - 5 u - 4 = 0

Solve with the quadratic formula

9 u^{2} - 5 u - 4 = 0

Quadratic Equation Formula:

For

a = 9, b = - 5, c = - 4

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 9 ( - 4 )}{2 \cdot 9}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 9 ( - 4 )}{2 \cdot 9}

(- 5)^{2} - 4 \cdot 9 (- 4) = 13

(- 5)^{2} - 4 \cdot 9 (- 4)

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 9 \cdot 4

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 9 \cdot 4

Multiply the numbers:

4 \cdot 9 \cdot 4 = 144

= 5^{2} + 144

5^{2} = 25

= 25 + 144

Add the numbers:

25 + 144 = 169

= 169

Factor the number:

169 = 1 3^{2}

= 1 3^{2}

Apply radical rule:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- ( - 5 ) \pm 13}{2 \cdot 9}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 13}{2 \cdot 9}, u_{2} = \frac{- ( - 5 ) - 13}{2 \cdot 9}

u = \frac{- ( - 5 ) + 13}{2 \cdot 9} : 1

\frac{- ( - 5 ) + 13}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{5 + 13}{2 \cdot 9}

Add the numbers:

5 + 13 = 18

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

Multiply the numbers:

2 \cdot 9 = 18

= \frac{18}{18}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 5 ) - 13}{2 \cdot 9} : - \frac{4}{9}

\frac{- ( - 5 ) - 13}{2 \cdot 9}

Apply rule

- (- a) = a

= \frac{5 - 13}{2 \cdot 9}

Subtract the numbers:

5 - 13 = - 8

= \frac{- 8}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{- 8}{18}

Apply the fraction rule:

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

= - \frac{8}{18}

Cancel the common factor:

2

= - \frac{4}{9}

The solutions to the quadratic equation are:

u = 1, u = - \frac{4}{9}

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - \frac{4}{9}

cos (x) = 1, cos (x) = - \frac{4}{9}

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

cos (x) = - \frac{4}{9} : x = arccos (- \frac{4}{9}) + 2 πn, x = - arccos (- \frac{4}{9}) + 2 πn

cos (x) = - \frac{4}{9}

Apply trig inverse properties

cos (x) = - \frac{4}{9}

General solutions for

cos (x) = - \frac{4}{9}

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

x = arccos (- \frac{4}{9}) + 2 πn, x = - arccos (- \frac{4}{9}) + 2 πn

x = arccos (- \frac{4}{9}) + 2 πn, x = - arccos (- \frac{4}{9}) + 2 πn

Combine all the solutions

x = 2 πn, x = arccos (- \frac{4}{9}) + 2 πn, x = - arccos (- \frac{4}{9}) + 2 πn

Show solutions in decimal form

x = 2 πn, x = 2.03135 \dots + 2 πn, x = - 2.03135 \dots + 2 πn

Popular Examples

cos(t)=-3/5

cos (t) = - \frac{3}{5}

2-sin(θ)=3sin(θ)

2 - sin (θ) = 3 sin (θ)

-cos(x/2)+sin(x/2)=sqrt(2/5)

- cos (\frac{x}{2}) + sin (\frac{x}{2}) = \frac{2}{5}

(35)/(sin(90))=(27)/(sin(y))

\frac{35}{sin ( 9 0 ^{\circ} )} = \frac{27}{sin ( y )}

4cos^2(x)+2cos(x)-2sqrt(2)cos(x)=sqrt(2)

4 cos^{2} (x) + 2 cos (x) - 22 cos (x) = 2

Frequently Asked Questions (FAQ)

What is the general solution for -9sin^2(x)-5cos(x)+5=0 ?
The general solution for -9sin^2(x)-5cos(x)+5=0 is x=2pin,x=2.03135…+2pin,x=-2.03135…+2pin