What is the general solution for 6cos^2(x)-7cos(x)-5=0 ?

The general solution for 6cos^2(x)-7cos(x)-5=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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6cos^2(x)-7cos(x)-5=0

Solution

6 cos^{2} (x) - 7 cos (x) - 5 = 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

6 cos^{2} (x) - 7 cos (x) - 5 = 0

Solve by substitution

6 cos^{2} (x) - 7 cos (x) - 5 = 0

Let:

cos (x) = u

6 u^{2} - 7 u - 5 = 0

6 u^{2} - 7 u - 5 = 0 : u = \frac{5}{3}, u = - \frac{1}{2}

6 u^{2} - 7 u - 5 = 0

Solve with the quadratic formula

6 u^{2} - 7 u - 5 = 0

Quadratic Equation Formula:

For

a = 6, b = - 7, c = - 5

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 ( - 5 )}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 ( - 5 )}{2 \cdot 6}

(- 7)^{2} - 4 \cdot 6 (- 5) = 13

(- 7)^{2} - 4 \cdot 6 (- 5)

Apply rule

- (- a) = a

= (- 7)^{2} + 4 \cdot 6 \cdot 5

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} + 4 \cdot 6 \cdot 5

Multiply the numbers:

4 \cdot 6 \cdot 5 = 120

= 7^{2} + 120

7^{2} = 49

= 49 + 120

Add the numbers:

49 + 120 = 169

= 169

Factor the number:

169 = 1 3^{2}

= 1 3^{2}

Apply radical rule:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- ( - 7 ) \pm 13}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 13}{2 \cdot 6}, u_{2} = \frac{- ( - 7 ) - 13}{2 \cdot 6}

u = \frac{- ( - 7 ) + 13}{2 \cdot 6} : \frac{5}{3}

\frac{- ( - 7 ) + 13}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{7 + 13}{2 \cdot 6}

Add the numbers:

7 + 13 = 20

= \frac{20}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{20}{12}

Cancel the common factor:

4

= \frac{5}{3}

u = \frac{- ( - 7 ) - 13}{2 \cdot 6} : - \frac{1}{2}

\frac{- ( - 7 ) - 13}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{7 - 13}{2 \cdot 6}

Subtract the numbers:

7 - 13 = - 6

= \frac{- 6}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6}{12}

Apply the fraction rule:

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

= - \frac{6}{12}

Cancel the common factor:

6

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{5}{3}, u = - \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = \frac{5}{3}, cos (x) = - \frac{1}{2}

cos (x) = \frac{5}{3}, cos (x) = - \frac{1}{2}

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

No Solution

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

- 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

cos(x)=(-3)/(5sin^2(x))

cos (x) = \frac{- 3}{5 sin ^{2} ( x )}

5cos^2(x)-12sin^2(x)=13

5 cos^{2} (x) - 12 sin^{2} (x) = 13

cos^2(x)-0.5cos(x)=0

cos^{2} (x) - 0.5 cos (x) = 0

5sin(x)cos(x)=2cos(x)

5 sin (x) cos (x) = 2 cos (x)

sin(x)=cos^3(x)

sin (x) = cos^{3} (x)

Frequently Asked Questions (FAQ)

What is the general solution for 6cos^2(x)-7cos(x)-5=0 ?
The general solution for 6cos^2(x)-7cos(x)-5=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin