What is the general solution for 4cos^2(x)-5cos(x)-6=0 ?

The general solution for 4cos^2(x)-5cos(x)-6=0 is x=2.41885…+2pin,x=-2.41885…+2pin

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4cos^2(x)-5cos(x)-6=0

Solution

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

Solution

x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn

Degrees

x = 138.59037 \dots^{\circ} + 36 0^{\circ} n, x = - 138.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

4 u^{2} - 5 u - 6 = 0 : u = 2, u = - \frac{3}{4}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = - 5, c = - 6

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

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

(- 5)^{2} - 4 \cdot 4 (- 6) = 11

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 6 = 96

= 5^{2} + 96

5^{2} = 25

= 25 + 96

Add the numbers:

25 + 96 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- ( - 5 ) \pm 11}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 11}{2 \cdot 4}, u_{2} = \frac{- ( - 5 ) - 11}{2 \cdot 4}

u = \frac{- ( - 5 ) + 11}{2 \cdot 4} : 2

\frac{- ( - 5 ) + 11}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{5 + 11}{2 \cdot 4}

Add the numbers:

5 + 11 = 16

= \frac{16}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= 2

u = \frac{- ( - 5 ) - 11}{2 \cdot 4} : - \frac{3}{4}

\frac{- ( - 5 ) - 11}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{5 - 11}{2 \cdot 4}

Subtract the numbers:

5 - 11 = - 6

= \frac{- 6}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 6}{8}

Apply the fraction rule:

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

= - \frac{6}{8}

Cancel the common factor:

2

= - \frac{3}{4}

The solutions to the quadratic equation are:

u = 2, u = - \frac{3}{4}

Substitute back

u = cos (x)

cos (x) = 2, cos (x) = - \frac{3}{4}

cos (x) = 2, cos (x) = - \frac{3}{4}

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

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

cos (x) = - \frac{3}{4}

Apply trig inverse properties

cos (x) = - \frac{3}{4}

General solutions for

cos (x) = - \frac{3}{4}

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

x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

Combine all the solutions

x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

Show solutions in decimal form

x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-5cos(x)-6=0 ?
The general solution for 4cos^2(x)-5cos(x)-6=0 is x=2.41885…+2pin,x=-2.41885…+2pin