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

The general solution for 2cos(2x)+3cos(x)-2=0 is x=0.80515…+2pin,x=2pi-0.80515…+2pin

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Popular >

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

Solution

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

Solution

x = 0.80515 \dots + 2 πn, x = 2 π - 0.80515 \dots + 2 πn

Degrees

x = 46.13190 \dots^{\circ} + 36 0^{\circ} n, x = 313.86809 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

- 2 + 2 cos (2 x) + 3 cos (x)

Use the Double Angle identity:

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

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

Simplify

- 2 + 2 (2 cos^{2} (x) - 1) + 3 cos (x) : 4 cos^{2} (x) + 3 cos (x) - 4

- 2 + 2 (2 cos^{2} (x) - 1) + 3 cos (x)

Expand

2 (2 cos^{2} (x) - 1) : 4 cos^{2} (x) - 2

2 (2 cos^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 2 cos^{2} (x), c = 1

= 2 \cdot 2 cos^{2} (x) - 2 \cdot 1

Simplify

2 \cdot 2 cos^{2} (x) - 2 \cdot 1 : 4 cos^{2} (x) - 2

2 \cdot 2 cos^{2} (x) - 2 \cdot 1

Multiply the numbers:

2 \cdot 2 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

= 4 cos^{2} (x) - 2

= 4 cos^{2} (x) - 2

= - 2 + 4 cos^{2} (x) - 2 + 3 cos (x)

Simplify

- 2 + 4 cos^{2} (x) - 2 + 3 cos (x) : 4 cos^{2} (x) + 3 cos (x) - 4

- 2 + 4 cos^{2} (x) - 2 + 3 cos (x)

Group like terms

= 4 cos^{2} (x) + 3 cos (x) - 2 - 2

Subtract the numbers:

- 2 - 2 = - 4

= 4 cos^{2} (x) + 3 cos (x) - 4

= 4 cos^{2} (x) + 3 cos (x) - 4

= 4 cos^{2} (x) + 3 cos (x) - 4

- 4 + 3 cos (x) + 4 cos^{2} (x) = 0

Solve by substitution

- 4 + 3 cos (x) + 4 cos^{2} (x) = 0

Let:

cos (x) = u

- 4 + 3 u + 4 u^{2} = 0

- 4 + 3 u + 4 u^{2} = 0 : u = \frac{- 3 + 73}{8}, u = \frac{- 3 - 73}{8}

- 4 + 3 u + 4 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

4 u^{2} + 3 u - 4 = 0

Solve with the quadratic formula

4 u^{2} + 3 u - 4 = 0

Quadratic Equation Formula:

For

a = 4, b = 3, c = - 4

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

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

3^{2} - 4 \cdot 4 (- 4) = 73

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 3^{2} + 64

3^{2} = 9

= 9 + 64

Add the numbers:

9 + 64 = 73

= 73

u_{1, 2} = \frac{- 3 \pm 73}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 3 + 73}{2 \cdot 4}, u_{2} = \frac{- 3 - 73}{2 \cdot 4}

u = \frac{- 3 + 73}{2 \cdot 4} : \frac{- 3 + 73}{8}

\frac{- 3 + 73}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 3 + 73}{8}

u = \frac{- 3 - 73}{2 \cdot 4} : \frac{- 3 - 73}{8}

\frac{- 3 - 73}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 3 - 73}{8}

The solutions to the quadratic equation are:

u = \frac{- 3 + 73}{8}, u = \frac{- 3 - 73}{8}

Substitute back

u = cos (x)

cos (x) = \frac{- 3 + 73}{8}, cos (x) = \frac{- 3 - 73}{8}

cos (x) = \frac{- 3 + 73}{8}, cos (x) = \frac{- 3 - 73}{8}

cos (x) = \frac{- 3 + 73}{8} : x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

cos (x) = \frac{- 3 + 73}{8}

Apply trig inverse properties

cos (x) = \frac{- 3 + 73}{8}

General solutions for

cos (x) = \frac{- 3 + 73}{8}

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

x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

cos (x) = \frac{- 3 - 73}{8} :

No Solution

cos (x) = \frac{- 3 - 73}{8}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (\frac{- 3 + 73}{8}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 73}{8}) + 2 πn

Show solutions in decimal form

x = 0.80515 \dots + 2 πn, x = 2 π - 0.80515 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2x)+3cos(x)-2=0 ?
The general solution for 2cos(2x)+3cos(x)-2=0 is x=0.80515…+2pin,x=2pi-0.80515…+2pin