What is the general solution for 3cos^2(x)-8cos(x)+4=0,0<= x<= 180 ?

The general solution for 3cos^2(x)-8cos(x)+4=0,0<= x<= 180 is x=0.84106…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3cos^2(x)-8cos(x)+4=0,0<= x<= 180

Solution

3 cos^{2} (x) - 8 cos (x) + 4 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

Solution

x = 0.84106 \dots

Solution steps

3 cos^{2} (x) - 8 cos (x) + 4 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

Solve by substitution

3 cos^{2} (x) - 8 cos (x) + 4 = 0

Let:

cos (x) = u

3 u^{2} - 8 u + 4 = 0

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

3 u^{2} - 8 u + 4 = 0

Solve with the quadratic formula

3 u^{2} - 8 u + 4 = 0

Quadratic Equation Formula:

For

a = 3, b = - 8, c = 4

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 3 \cdot 4}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 3 \cdot 4}{2 \cdot 3}

(- 8)^{2} - 4 \cdot 3 \cdot 4 = 4

(- 8)^{2} - 4 \cdot 3 \cdot 4

Apply exponent rule:

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

n

is even

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

= 8^{2} - 4 \cdot 3 \cdot 4

Multiply the numbers:

4 \cdot 3 \cdot 4 = 48

= 8^{2} - 48

8^{2} = 64

= 64 - 48

Subtract the numbers:

64 - 48 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

u_{1, 2} = \frac{- ( - 8 ) \pm 4}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 4}{2 \cdot 3}, u_{2} = \frac{- ( - 8 ) - 4}{2 \cdot 3}

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

\frac{- ( - 8 ) + 4}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{8 + 4}{2 \cdot 3}

Add the numbers:

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

\frac{- ( - 8 ) - 4}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{8 - 4}{2 \cdot 3}

Subtract the numbers:

8 - 4 = 4

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = 2, 0 \leq x \leq 18 0^{\circ} :

No Solution

cos (x) = 2, 0 \leq x \leq 18 0^{\circ}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{2}{3}, 0 \leq x \leq 18 0^{\circ} : x = arccos (\frac{2}{3})

cos (x) = \frac{2}{3}, 0 \leq x \leq 18 0^{\circ}

Apply trig inverse properties

cos (x) = \frac{2}{3}

General solutions for

cos (x) = \frac{2}{3}

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

x = arccos (\frac{2}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{2}{3}) + 36 0^{\circ} n

x = arccos (\frac{2}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{2}{3}) + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 18 0^{\circ}

x = arccos (\frac{2}{3})

Combine all the solutions

x = arccos (\frac{2}{3})

Show solutions in decimal form

x = 0.84106 \dots

Popular Examples

sin(x)= 5/25

sin (x) = \frac{5}{25}

sin(x)cot(x)=-(sqrt(2))/2

sin (x) cot (x) = - \frac{2}{2}

3tan^2(θ)-1=sec^2(θ)

3 tan^{2} (θ) - 1 = sec^{2} (θ)

75cos^2(x)-34cos(x)+3=0

75 cos^{2} (x) - 34 cos (x) + 3 = 0

cos(2x)-sin(x)=0,0<= x<= 2pi

cos (2 x) - sin (x) = 0, 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)-8cos(x)+4=0,0<= x<= 180 ?
The general solution for 3cos^2(x)-8cos(x)+4=0,0<= x<= 180 is x=0.84106…