What is the general solution for 10cos^2(x)+5cos(x)-5=0 ?

The general solution for 10cos^2(x)+5cos(x)-5=0 is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin

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10cos^2(x)+5cos(x)-5=0

Solution

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

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = π + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

10 u^{2} + 5 u - 5 = 0

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

10 u^{2} + 5 u - 5 = 0

Solve with the quadratic formula

10 u^{2} + 5 u - 5 = 0

Quadratic Equation Formula:

For

a = 10, b = 5, c = - 5

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 10 ( - 5 )}{2 \cdot 10}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 10 ( - 5 )}{2 \cdot 10}

5^{2} - 4 \cdot 10 (- 5) = 15

5^{2} - 4 \cdot 10 (- 5)

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 10 \cdot 5

Multiply the numbers:

4 \cdot 10 \cdot 5 = 200

= 5^{2} + 200

5^{2} = 25

= 25 + 200

Add the numbers:

25 + 200 = 225

= 225

Factor the number:

225 = 1 5^{2}

= 1 5^{2}

Apply radical rule:

1 5^{2} = 15

= 15

u_{1, 2} = \frac{- 5 \pm 15}{2 \cdot 10}

Separate the solutions

u_{1} = \frac{- 5 + 15}{2 \cdot 10}, u_{2} = \frac{- 5 - 15}{2 \cdot 10}

u = \frac{- 5 + 15}{2 \cdot 10} : \frac{1}{2}

\frac{- 5 + 15}{2 \cdot 10}

Add/Subtract the numbers:

- 5 + 15 = 10

= \frac{10}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{10}{20}

Cancel the common factor:

10

= \frac{1}{2}

u = \frac{- 5 - 15}{2 \cdot 10} : - 1

\frac{- 5 - 15}{2 \cdot 10}

Subtract the numbers:

- 5 - 15 = - 20

= \frac{- 20}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 20}{20}

Apply the fraction rule:

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

= - \frac{20}{20}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - 1

Substitute back

u = cos (x)

cos (x) = \frac{1}{2}, cos (x) = - 1

cos (x) = \frac{1}{2}, cos (x) = - 1

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

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 = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 10cos^2(x)+5cos(x)-5=0 ?
The general solution for 10cos^2(x)+5cos(x)-5=0 is x= pi/3+2pin,x=(5pi)/3+2pin,x=pi+2pin