What is the general solution for 4cos^2(x)-18cos(x)-10=0 ?

The general solution for 4cos^2(x)-18cos(x)-10=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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4cos^2(x)-18cos(x)-10=0

Solution

4 cos^{2} (x) - 18 cos (x) - 10 = 0

Solution

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (x) - 18 cos (x) - 10 = 0

Solve by substitution

4 cos^{2} (x) - 18 cos (x) - 10 = 0

Let:

cos (x) = u

4 u^{2} - 18 u - 10 = 0

4 u^{2} - 18 u - 10 = 0 : u = 5, u = - \frac{1}{2}

4 u^{2} - 18 u - 10 = 0

Solve with the quadratic formula

4 u^{2} - 18 u - 10 = 0

Quadratic Equation Formula:

For

a = 4, b = - 18, c = - 10

u_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 4 ( - 10 )}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 4 ( - 10 )}{2 \cdot 4}

(- 18)^{2} - 4 \cdot 4 (- 10) = 22

(- 18)^{2} - 4 \cdot 4 (- 10)

Apply rule

- (- a) = a

= (- 18)^{2} + 4 \cdot 4 \cdot 10

Apply exponent rule:

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

n

is even

(- 18)^{2} = 1 8^{2}

= 1 8^{2} + 4 \cdot 4 \cdot 10

Multiply the numbers:

4 \cdot 4 \cdot 10 = 160

= 1 8^{2} + 160

1 8^{2} = 324

= 324 + 160

Add the numbers:

324 + 160 = 484

= 484

Factor the number:

484 = 2 2^{2}

= 2 2^{2}

Apply radical rule:

2 2^{2} = 22

= 22

u_{1, 2} = \frac{- ( - 18 ) \pm 22}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 18 ) + 22}{2 \cdot 4}, u_{2} = \frac{- ( - 18 ) - 22}{2 \cdot 4}

u = \frac{- ( - 18 ) + 22}{2 \cdot 4} : 5

\frac{- ( - 18 ) + 22}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{18 + 22}{2 \cdot 4}

Add the numbers:

18 + 22 = 40

= \frac{40}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{40}{8}

Divide the numbers:

\frac{40}{8} = 5

= 5

u = \frac{- ( - 18 ) - 22}{2 \cdot 4} : - \frac{1}{2}

\frac{- ( - 18 ) - 22}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{18 - 22}{2 \cdot 4}

Subtract the numbers:

18 - 22 = - 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = 5, u = - \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = 5, cos (x) = - \frac{1}{2}

cos (x) = 5, cos (x) = - \frac{1}{2}

cos (x) = 5 :

No Solution

cos (x) = 5

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Popular Examples

25sin(θ)-1.5cos(θ)=20

25 sin (θ) - 1.5 cos (θ) = 20

sin((2θ)/3)= 1/2 ,0<= θ<= 2pi

sin (\frac{2 θ}{3}) = \frac{1}{2}, 0 \leq θ \leq 2 π

sqrt(1-sin^2(x))=e^{1/2 ln(1-cos^2(x))}

1 - sin^{2} (x) = e^{\frac{1}{2} l n (1 - c o s^{2} (x))}

cot^2(x)=tan(1/2 x)

cot^{2} (x) = tan (\frac{1}{2} x)

cos(C)= 3/4

cos (C) = \frac{3}{4}

Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-18cos(x)-10=0 ?
The general solution for 4cos^2(x)-18cos(x)-10=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin