What is the general solution for -10cos(2x)-32cos(x)-22=0 ?

The general solution for -10cos(2x)-32cos(x)-22=0 is x=pi+2pin,x=2.21429…+2pin,x=-2.21429…+2pin

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-10cos(2x)-32cos(x)-22=0

Solution

- 10 cos (2 x) - 32 cos (x) - 22 = 0

Solution

x = π + 2 πn, x = 2.21429 \dots + 2 πn, x = - 2.21429 \dots + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n, x = 126.86989 \dots^{\circ} + 36 0^{\circ} n, x = - 126.86989 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 10 cos (2 x) - 32 cos (x) - 22 = 0

Rewrite using trig identities

- 22 - 10 cos (2 x) - 32 cos (x)

Use the Double Angle identity:

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

= - 22 - 10 (2 cos^{2} (x) - 1) - 32 cos (x)

Simplify

- 22 - 10 (2 cos^{2} (x) - 1) - 32 cos (x) : - 20 cos^{2} (x) - 32 cos (x) - 12

- 22 - 10 (2 cos^{2} (x) - 1) - 32 cos (x)

Expand

- 10 (2 cos^{2} (x) - 1) : - 20 cos^{2} (x) + 10

- 10 (2 cos^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = - 10, b = 2 cos^{2} (x), c = 1

= - 10 \cdot 2 cos^{2} (x) - (- 10) \cdot 1

Apply minus-plus rules

- (- a) = a

= - 10 \cdot 2 cos^{2} (x) + 10 \cdot 1

Simplify

- 10 \cdot 2 cos^{2} (x) + 10 \cdot 1 : - 20 cos^{2} (x) + 10

- 10 \cdot 2 cos^{2} (x) + 10 \cdot 1

Multiply the numbers:

10 \cdot 2 = 20

= - 20 cos^{2} (x) + 10 \cdot 1

Multiply the numbers:

10 \cdot 1 = 10

= - 20 cos^{2} (x) + 10

= - 20 cos^{2} (x) + 10

= - 22 - 20 cos^{2} (x) + 10 - 32 cos (x)

Simplify

- 22 - 20 cos^{2} (x) + 10 - 32 cos (x) : - 20 cos^{2} (x) - 32 cos (x) - 12

- 22 - 20 cos^{2} (x) + 10 - 32 cos (x)

Group like terms

= - 20 cos^{2} (x) - 32 cos (x) - 22 + 10

Add/Subtract the numbers:

- 22 + 10 = - 12

= - 20 cos^{2} (x) - 32 cos (x) - 12

= - 20 cos^{2} (x) - 32 cos (x) - 12

= - 20 cos^{2} (x) - 32 cos (x) - 12

- 12 - 20 cos^{2} (x) - 32 cos (x) = 0

Solve by substitution

- 12 - 20 cos^{2} (x) - 32 cos (x) = 0

Let:

cos (x) = u

- 12 - 20 u^{2} - 32 u = 0

- 12 - 20 u^{2} - 32 u = 0 : u = - 1, u = - \frac{3}{5}

- 12 - 20 u^{2} - 32 u = 0

Write in the standard form

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

- 20 u^{2} - 32 u - 12 = 0

Solve with the quadratic formula

- 20 u^{2} - 32 u - 12 = 0

Quadratic Equation Formula:

For

a = - 20, b = - 32, c = - 12

u_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 ( - 20 ) ( - 12 )}{2 ( - 20 )}

u_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 ( - 20 ) ( - 12 )}{2 ( - 20 )}

(- 32)^{2} - 4 (- 20) (- 12) = 8

(- 32)^{2} - 4 (- 20) (- 12)

Apply rule

- (- a) = a

= (- 32)^{2} - 4 \cdot 20 \cdot 12

Apply exponent rule:

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

n

is even

(- 32)^{2} = 3 2^{2}

= 3 2^{2} - 4 \cdot 20 \cdot 12

Multiply the numbers:

4 \cdot 20 \cdot 12 = 960

= 3 2^{2} - 960

3 2^{2} = 1024

= 1024 - 960

Subtract the numbers:

1024 - 960 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

n a^{n} = a

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 32 ) \pm 8}{2 ( - 20 )}

Separate the solutions

u_{1} = \frac{- ( - 32 ) + 8}{2 ( - 20 )}, u_{2} = \frac{- ( - 32 ) - 8}{2 ( - 20 )}

u = \frac{- ( - 32 ) + 8}{2 ( - 20 )} : - 1

\frac{- ( - 32 ) + 8}{2 ( - 20 )}

Remove parentheses:

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

= \frac{32 + 8}{- 2 \cdot 20}

Add the numbers:

32 + 8 = 40

= \frac{40}{- 2 \cdot 20}

Multiply the numbers:

2 \cdot 20 = 40

= \frac{40}{- 40}

Apply the fraction rule:

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

= - \frac{40}{40}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 32 ) - 8}{2 ( - 20 )} : - \frac{3}{5}

\frac{- ( - 32 ) - 8}{2 ( - 20 )}

Remove parentheses:

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

= \frac{32 - 8}{- 2 \cdot 20}

Subtract the numbers:

32 - 8 = 24

= \frac{24}{- 2 \cdot 20}

Multiply the numbers:

2 \cdot 20 = 40

= \frac{24}{- 40}

Apply the fraction rule:

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

= - \frac{24}{40}

Cancel the common factor:

8

= - \frac{3}{5}

The solutions to the quadratic equation are:

u = - 1, u = - \frac{3}{5}

Substitute back

u = cos (x)

cos (x) = - 1, cos (x) = - \frac{3}{5}

cos (x) = - 1, cos (x) = - \frac{3}{5}

cos (x) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

cos (x) = - \frac{3}{5} : x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

Combine all the solutions

x = π + 2 πn, x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

Show solutions in decimal form

x = π + 2 πn, x = 2.21429 \dots + 2 πn, x = - 2.21429 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for -10cos(2x)-32cos(x)-22=0 ?
The general solution for -10cos(2x)-32cos(x)-22=0 is x=pi+2pin,x=2.21429…+2pin,x=-2.21429…+2pin