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

The general solution for cos^2(x)+5cos(x)-3=0 is x=0.99871…+2pin,x=2pi-0.99871…+2pin

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cos^2(x)+5cos(x)-3=0

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

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) + 5 cos (x) - 3 = 0

Solve by substitution

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

Let:

cos (x) = u

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

u^{2} + 5 u - 3 = 0 : u = \frac{- 5 + 37}{2}, u = \frac{- 5 - 37}{2}

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

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22 sin (x) + sec (x) = 0

cos (θ) = \frac{- 2}{( 30 + ( 7 ) )}

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What is the general solution for cos^2(x)+5cos(x)-3=0 ?
The general solution for cos^2(x)+5cos(x)-3=0 is x=0.99871…+2pin,x=2pi-0.99871…+2pin