What is the general solution for 4cos^2(x)-3cos(x)-2=0 ?

The general solution for 4cos^2(x)-3cos(x)-2=0 is x=2.01018…+2pin,x=-2.01018…+2pin

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4cos^2(x)-3cos(x)-2=0

Solution

4 cos^{2} (x) - 3 cos (x) - 2 = 0

Solution

x = 2.01018 \dots + 2 πn, x = - 2.01018 \dots + 2 πn

Degrees

x = 115.17538 \dots^{\circ} + 36 0^{\circ} n, x = - 115.17538 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (x) - 3 cos (x) - 2 = 0

Solve by substitution

4 cos^{2} (x) - 3 cos (x) - 2 = 0

Let:

cos (x) = u

4 u^{2} - 3 u - 2 = 0

4 u^{2} - 3 u - 2 = 0 : u = \frac{3 + 41}{8}, u = \frac{3 - 41}{8}

4 u^{2} - 3 u - 2 = 0

Solve with the quadratic formula

4 u^{2} - 3 u - 2 = 0

Quadratic Equation Formula:

For

a = 4, b = - 3, c = - 2

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 ( - 2 )}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 ( - 2 )}{2 \cdot 4}

(- 3)^{2} - 4 \cdot 4 (- 2) = 41

(- 3)^{2} - 4 \cdot 4 (- 2)

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 4 \cdot 2

Apply exponent rule:

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

n

is even

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

= 3^{2} + 4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 3^{2} + 32

3^{2} = 9

= 9 + 32

Add the numbers:

9 + 32 = 41

= 41

u_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 41}{2 \cdot 4}, u_{2} = \frac{- ( - 3 ) - 41}{2 \cdot 4}

u = \frac{- ( - 3 ) + 41}{2 \cdot 4} : \frac{3 + 41}{8}

\frac{- ( - 3 ) + 41}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{3 + 41}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{3 + 41}{8}

u = \frac{- ( - 3 ) - 41}{2 \cdot 4} : \frac{3 - 41}{8}

\frac{- ( - 3 ) - 41}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{3 - 41}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{3 - 41}{8}

The solutions to the quadratic equation are:

u = \frac{3 + 41}{8}, u = \frac{3 - 41}{8}

Substitute back

u = cos (x)

cos (x) = \frac{3 + 41}{8}, cos (x) = \frac{3 - 41}{8}

cos (x) = \frac{3 + 41}{8}, cos (x) = \frac{3 - 41}{8}

cos (x) = \frac{3 + 41}{8} :

No Solution

cos (x) = \frac{3 + 41}{8}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{3 - 41}{8} : x = arccos (\frac{3 - 41}{8}) + 2 πn, x = - arccos (\frac{3 - 41}{8}) + 2 πn

cos (x) = \frac{3 - 41}{8}

Apply trig inverse properties

cos (x) = \frac{3 - 41}{8}

General solutions for

cos (x) = \frac{3 - 41}{8}

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

x = arccos (\frac{3 - 41}{8}) + 2 πn, x = - arccos (\frac{3 - 41}{8}) + 2 πn

x = arccos (\frac{3 - 41}{8}) + 2 πn, x = - arccos (\frac{3 - 41}{8}) + 2 πn

Combine all the solutions

x = arccos (\frac{3 - 41}{8}) + 2 πn, x = - arccos (\frac{3 - 41}{8}) + 2 πn

Show solutions in decimal form

x = 2.01018 \dots + 2 πn, x = - 2.01018 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-3cos(x)-2=0 ?
The general solution for 4cos^2(x)-3cos(x)-2=0 is x=2.01018…+2pin,x=-2.01018…+2pin