What is the general solution for 3cos^2(x)-10cos(x)+3=0 ?

The general solution for 3cos^2(x)-10cos(x)+3=0 is x=1.23095…+2pin,x=2pi-1.23095…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3cos^2(x)-10cos(x)+3=0

Solution

3 cos^{2} (x) - 10 cos (x) + 3 = 0

Solution

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

Degrees

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 cos^{2} (x) - 10 cos (x) + 3 = 0

Solve by substitution

3 cos^{2} (x) - 10 cos (x) + 3 = 0

Let:

cos (x) = u

3 u^{2} - 10 u + 3 = 0

3 u^{2} - 10 u + 3 = 0 : u = 3, u = \frac{1}{3}

3 u^{2} - 10 u + 3 = 0

Solve with the quadratic formula

3 u^{2} - 10 u + 3 = 0

Quadratic Equation Formula:

For

a = 3, b = - 10, c = 3

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

(- 10)^{2} - 4 \cdot 3 \cdot 3 = 8

(- 10)^{2} - 4 \cdot 3 \cdot 3

Apply exponent rule:

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

n

is even

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 3 \cdot 3

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 1 0^{2} - 36

1 0^{2} = 100

= 100 - 36

Subtract the numbers:

100 - 36 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 10 ) \pm 8}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 10 ) + 8}{2 \cdot 3}, u_{2} = \frac{- ( - 10 ) - 8}{2 \cdot 3}

u = \frac{- ( - 10 ) + 8}{2 \cdot 3} : 3

\frac{- ( - 10 ) + 8}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{10 + 8}{2 \cdot 3}

Add the numbers:

10 + 8 = 18

= \frac{18}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{18}{6}

Divide the numbers:

\frac{18}{6} = 3

= 3

u = \frac{- ( - 10 ) - 8}{2 \cdot 3} : \frac{1}{3}

\frac{- ( - 10 ) - 8}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{10 - 8}{2 \cdot 3}

Subtract the numbers:

10 - 8 = 2

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{2}{6}

Cancel the common factor:

2

= \frac{1}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = 3 :

No Solution

cos (x) = 3

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{1}{3} : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

Combine all the solutions

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

Show solutions in decimal form

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

Popular Examples

(m+1)sin(x)+2-m=0

(m + 1) sin (x) + 2 - m = 0

sin^5(x)+sin(x)+2sin^2(x)=1

sin^{5} (x) + sin (x) + 2 sin^{2} (x) = 1

(2cos(x)-1)*(2sin(x)+1)=3-4sin^2(x)

(2 cos (x) - 1) \cdot (2 sin (x) + 1) = 3 - 4 sin^{2} (x)

r=(2a)/((1+cos(x)))

r = \frac{2 a}{( 1 + cos ( x ) )}

cos^3(x)+cos^2(x)+cos(x)+1=0

cos^{3} (x) + cos^{2} (x) + cos (x) + 1 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)-10cos(x)+3=0 ?
The general solution for 3cos^2(x)-10cos(x)+3=0 is x=1.23095…+2pin,x=2pi-1.23095…+2pin