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

The general solution for cos^2(x)=3-5cos(x) is x=0.99871…+2pin,x=2pi-0.99871…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos^2(x)=3-5cos(x)

Solution

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

Solution

x = 0.99871 \dots + 2 πn, x = 2 π - 0.99871 \dots + 2 πn

Degrees

x = 57.22228 \dots^{\circ} + 36 0^{\circ} n, x = 302.77771 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

u^{2} = 3 - 5 u

u^{2} = 3 - 5 u : u = \frac{- 5 + 37}{2}, u = \frac{- 5 - 37}{2}

u^{2} = 3 - 5 u

Move

5 u

to the left side

u^{2} = 3 - 5 u

Add

5 u

to both sides

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

Simplify

u^{2} + 5 u = 3

u^{2} + 5 u = 3

Move

3

to the left side

u^{2} + 5 u = 3

Subtract

3

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 5, c = - 3

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

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

5^{2} - 4 \cdot 1 \cdot (- 3) = 37

5^{2} - 4 \cdot 1 \cdot (- 3)

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 5^{2} + 12

5^{2} = 25

= 25 + 12

Add the numbers:

25 + 12 = 37

= 37

u_{1, 2} = \frac{- 5 \pm 37}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 5 + 37}{2 \cdot 1}, u_{2} = \frac{- 5 - 37}{2 \cdot 1}

u = \frac{- 5 + 37}{2 \cdot 1} : \frac{- 5 + 37}{2}

\frac{- 5 + 37}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 + 37}{2}

u = \frac{- 5 - 37}{2 \cdot 1} : \frac{- 5 - 37}{2}

\frac{- 5 - 37}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 - 37}{2}

The solutions to the quadratic equation are:

u = \frac{- 5 + 37}{2}, u = \frac{- 5 - 37}{2}

Substitute back

u = cos (x)

cos (x) = \frac{- 5 + 37}{2}, cos (x) = \frac{- 5 - 37}{2}

cos (x) = \frac{- 5 + 37}{2}, cos (x) = \frac{- 5 - 37}{2}

cos (x) = \frac{- 5 + 37}{2} : x = arccos (\frac{- 5 + 37}{2}) + 2 πn, x = 2 π - arccos (\frac{- 5 + 37}{2}) + 2 πn

cos (x) = \frac{- 5 + 37}{2}

Apply trig inverse properties

cos (x) = \frac{- 5 + 37}{2}

General solutions for

cos (x) = \frac{- 5 + 37}{2}

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

x = arccos (\frac{- 5 + 37}{2}) + 2 πn, x = 2 π - arccos (\frac{- 5 + 37}{2}) + 2 πn

x = arccos (\frac{- 5 + 37}{2}) + 2 πn, x = 2 π - arccos (\frac{- 5 + 37}{2}) + 2 πn

cos (x) = \frac{- 5 - 37}{2} :

No Solution

cos (x) = \frac{- 5 - 37}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (\frac{- 5 + 37}{2}) + 2 πn, x = 2 π - arccos (\frac{- 5 + 37}{2}) + 2 πn

Show solutions in decimal form

x = 0.99871 \dots + 2 πn, x = 2 π - 0.99871 \dots + 2 πn

Popular Examples

sin(x)= 1/(sec(x))

sin (x) = \frac{1}{sec ( x )}

solvefor x,1-2cos^2(x)-2cos(x)=0 solve for

x, 1 - 2 cos^{2} (x) - 2 cos (x) = 0

1=tan(x/3)

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

cos(12x)=0

cos (12 x) = 0

cos(2x+1)= 1/2

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

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)=3-5cos(x) ?
The general solution for cos^2(x)=3-5cos(x) is x=0.99871…+2pin,x=2pi-0.99871…+2pin