What is the general solution for 3cos^2(x)-5cos(x)-2=0,(0,360) ?

The general solution for 3cos^2(x)-5cos(x)-2=0,(0,360) is x=1.91063…,x=-1.91063…+360

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3cos^2(x)-5cos(x)-2=0,(0,360)

Solution

3 cos^{2} (x) - 5 cos (x) - 2 = 0, (0^{\circ}, 36 0^{\circ})

Solution

x = 1.91063 \dots, x = - 1.91063 \dots + 36 0^{\circ}

Radians

x = 1.91063 \dots, x = - 1.91063 \dots + 2 π

Solution steps

3 cos^{2} (x) - 5 cos (x) - 2 = 0, (0^{\circ}, 36 0^{\circ})

Solve by substitution

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

Let:

cos (x) = u

3 u^{2} - 5 u - 2 = 0

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

3 u^{2} - 5 u - 2 = 0

Solve with the quadratic formula

3 u^{2} - 5 u - 2 = 0

Quadratic Equation Formula:

For

a = 3, b = - 5, c = - 2

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 (- 2) = 7

(- 5)^{2} - 4 \cdot 3 (- 2)

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 3 \cdot 2

Apply exponent rule:

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

n

is even

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

= 5^{2} + 4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 5 ) \pm 7}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 7}{2 \cdot 3}, u_{2} = \frac{- ( - 5 ) - 7}{2 \cdot 3}

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

\frac{- ( - 5 ) + 7}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{5 + 7}{2 \cdot 3}

Add the numbers:

5 + 7 = 12

= \frac{12}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{12}{6}

Divide the numbers:

\frac{12}{6} = 2

= 2

u = \frac{- ( - 5 ) - 7}{2 \cdot 3} : - \frac{1}{3}

\frac{- ( - 5 ) - 7}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{5 - 7}{2 \cdot 3}

Subtract the numbers:

5 - 7 = - 2

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2}{6}

Apply the fraction rule:

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

= - \frac{2}{6}

Cancel the common factor:

2

= - \frac{1}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = 2, (0, 36 0^{\circ}) :

No Solution

cos (x) = 2, (0, 36 0^{\circ})

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - \frac{1}{3}, (0, 36 0^{\circ}) : x = arccos (- \frac{1}{3}), x = - arccos (- \frac{1}{3}) + 36 0^{\circ}

cos (x) = - \frac{1}{3}, (0, 36 0^{\circ})

Apply trig inverse properties

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

General solutions for

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

cos (x) = - a \Rightarrow x = arccos (- a) + 36 0^{\circ} n, x = - arccos (- a) + 36 0^{\circ} n

x = arccos (- \frac{1}{3}) + 36 0^{\circ} n, x = - arccos (- \frac{1}{3}) + 36 0^{\circ} n

x = arccos (- \frac{1}{3}) + 36 0^{\circ} n, x = - arccos (- \frac{1}{3}) + 36 0^{\circ} n

Solutions for the range

(0, 36 0^{\circ})

x = arccos (- \frac{1}{3}), x = - arccos (- \frac{1}{3}) + 36 0^{\circ}

Combine all the solutions

x = arccos (- \frac{1}{3}), x = - arccos (- \frac{1}{3}) + 36 0^{\circ}

Show solutions in decimal form

x = 1.91063 \dots, x = - 1.91063 \dots + 36 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)-5cos(x)-2=0,(0,360) ?
The general solution for 3cos^2(x)-5cos(x)-2=0,(0,360) is x=1.91063…,x=-1.91063…+360