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

The general solution for cos^2(x)+5cos(x)-2=0 is x=1.18933…+2pin,x=2pi-1.18933…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos^2(x)+5cos(x)-2=0

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

x = 1.18933 \dots + 2 πn, x = 2 π - 1.18933 \dots + 2 πn

Degrees

x = 68.14361 \dots^{\circ} + 36 0^{\circ} n, x = 291.85638 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

u^{2} + 5 u - 2 = 0 : u = \frac{- 5 + 33}{2}, u = \frac{- 5 - 33}{2}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 5, c = - 2

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

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

5^{2} - 4 \cdot 1 \cdot (- 2) = 33

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 5^{2} + 8

5^{2} = 25

= 25 + 8

Add the numbers:

25 + 8 = 33

= 33

u_{1, 2} = \frac{- 5 \pm 33}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 5 + 33}{2 \cdot 1}, u_{2} = \frac{- 5 - 33}{2 \cdot 1}

u = \frac{- 5 + 33}{2 \cdot 1} : \frac{- 5 + 33}{2}

\frac{- 5 + 33}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 + 33}{2}

u = \frac{- 5 - 33}{2 \cdot 1} : \frac{- 5 - 33}{2}

\frac{- 5 - 33}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 5 - 33}{2}

The solutions to the quadratic equation are:

u = \frac{- 5 + 33}{2}, u = \frac{- 5 - 33}{2}

Substitute back

u = cos (x)

cos (x) = \frac{- 5 + 33}{2}, cos (x) = \frac{- 5 - 33}{2}

cos (x) = \frac{- 5 + 33}{2}, cos (x) = \frac{- 5 - 33}{2}

cos (x) = \frac{- 5 + 33}{2} : x = arccos (\frac{- 5 + 33}{2}) + 2 πn, x = 2 π - arccos (\frac{- 5 + 33}{2}) + 2 πn

cos (x) = \frac{- 5 + 33}{2}

Apply trig inverse properties

cos (x) = \frac{- 5 + 33}{2}

General solutions for

cos (x) = \frac{- 5 + 33}{2}

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

x = arccos (\frac{- 5 + 33}{2}) + 2 πn, x = 2 π - arccos (\frac{- 5 + 33}{2}) + 2 πn

x = arccos (\frac{- 5 + 33}{2}) + 2 πn, x = 2 π - arccos (\frac{- 5 + 33}{2}) + 2 πn

cos (x) = \frac{- 5 - 33}{2} :

No Solution

cos (x) = \frac{- 5 - 33}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (\frac{- 5 + 33}{2}) + 2 πn, x = 2 π - arccos (\frac{- 5 + 33}{2}) + 2 πn

Show solutions in decimal form

x = 1.18933 \dots + 2 πn, x = 2 π - 1.18933 \dots + 2 πn

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

(cos^{2} (x) + cos (x)) \cdot (sin (x) + sin^{3} (x)) = 0

arccos (x) - arcsin (x) = arcsin (1 - x)

sin^{2} (x) = ∣ sin (x) ∣

1 - cos^{2} (x) - sin^{22} (x) = 0

What is the general solution for cos^2(x)+5cos(x)-2=0 ?
The general solution for cos^2(x)+5cos(x)-2=0 is x=1.18933…+2pin,x=2pi-1.18933…+2pin