What is the general solution for 2cos^2(x)+9cos(x)=4 ?

The general solution for 2cos^2(x)+9cos(x)=4 is x=1.15104…+2pin,x=2pi-1.15104…+2pin

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Popular >

2cos^2(x)+9cos(x)=4

Solution

2 cos^{2} (x) + 9 cos (x) = 4

Solution

x = 1.15104 \dots + 2 πn, x = 2 π - 1.15104 \dots + 2 πn

Degrees

x = 65.94982 \dots^{\circ} + 36 0^{\circ} n, x = 294.05017 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (x) + 9 cos (x) = 4

Solve by substitution

2 cos^{2} (x) + 9 cos (x) = 4

Let:

cos (x) = u

2 u^{2} + 9 u = 4

2 u^{2} + 9 u = 4 : u = \frac{- 9 + 113}{4}, u = \frac{- 9 - 113}{4}

2 u^{2} + 9 u = 4

Move

4

to the left side

2 u^{2} + 9 u = 4

Subtract

4

from both sides

2 u^{2} + 9 u - 4 = 4 - 4

Simplify

2 u^{2} + 9 u - 4 = 0

2 u^{2} + 9 u - 4 = 0

Solve with the quadratic formula

2 u^{2} + 9 u - 4 = 0

Quadratic Equation Formula:

For

a = 2, b = 9, c = - 4

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

9^{2} - 4 \cdot 2 (- 4) = 113

9^{2} - 4 \cdot 2 (- 4)

Apply rule

- (- a) = a

= 9^{2} + 4 \cdot 2 \cdot 4

Multiply the numbers:

4 \cdot 2 \cdot 4 = 32

= 9^{2} + 32

9^{2} = 81

= 81 + 32

Add the numbers:

81 + 32 = 113

= 113

u_{1, 2} = \frac{- 9 \pm 113}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 9 + 113}{2 \cdot 2}, u_{2} = \frac{- 9 - 113}{2 \cdot 2}

u = \frac{- 9 + 113}{2 \cdot 2} : \frac{- 9 + 113}{4}

\frac{- 9 + 113}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 9 + 113}{4}

u = \frac{- 9 - 113}{2 \cdot 2} : \frac{- 9 - 113}{4}

\frac{- 9 - 113}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 9 - 113}{4}

The solutions to the quadratic equation are:

u = \frac{- 9 + 113}{4}, u = \frac{- 9 - 113}{4}

Substitute back

u = cos (x)

cos (x) = \frac{- 9 + 113}{4}, cos (x) = \frac{- 9 - 113}{4}

cos (x) = \frac{- 9 + 113}{4}, cos (x) = \frac{- 9 - 113}{4}

cos (x) = \frac{- 9 + 113}{4} : x = arccos (\frac{- 9 + 113}{4}) + 2 πn, x = 2 π - arccos (\frac{- 9 + 113}{4}) + 2 πn

cos (x) = \frac{- 9 + 113}{4}

Apply trig inverse properties

cos (x) = \frac{- 9 + 113}{4}

General solutions for

cos (x) = \frac{- 9 + 113}{4}

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

x = arccos (\frac{- 9 + 113}{4}) + 2 πn, x = 2 π - arccos (\frac{- 9 + 113}{4}) + 2 πn

x = arccos (\frac{- 9 + 113}{4}) + 2 πn, x = 2 π - arccos (\frac{- 9 + 113}{4}) + 2 πn

cos (x) = \frac{- 9 - 113}{4} :

No Solution

cos (x) = \frac{- 9 - 113}{4}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (\frac{- 9 + 113}{4}) + 2 πn, x = 2 π - arccos (\frac{- 9 + 113}{4}) + 2 πn

Show solutions in decimal form

x = 1.15104 \dots + 2 πn, x = 2 π - 1.15104 \dots + 2 πn

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sin (2 x + 60) = \frac{1}{2}, - 18 0^{\circ} \leq, x \leq 18 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)+9cos(x)=4 ?
The general solution for 2cos^2(x)+9cos(x)=4 is x=1.15104…+2pin,x=2pi-1.15104…+2pin