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Popular >

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

Solution

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

Solution

x = 0.89095 \dots + 2 πn, x = 2 π - 0.89095 \dots + 2 πn, x = 2.49035 \dots + 2 πn, x = - 2.49035 \dots + 2 πn

Degrees

x = 51.04815 \dots^{\circ} + 36 0^{\circ} n, x = 308.95184 \dots^{\circ} + 36 0^{\circ} n, x = 142.68678 \dots^{\circ} + 36 0^{\circ} n, x = - 142.68678 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

cos (x) + 3 cos (2 x)

Use the Double Angle identity:

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

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

cos (x) + (- 1 + 2 cos^{2} (x)) \cdot 3 = 0

Solve by substitution

cos (x) + (- 1 + 2 cos^{2} (x)) \cdot 3 = 0

Let:

cos (x) = u

u + (- 1 + 2 u^{2}) \cdot 3 = 0

u + (- 1 + 2 u^{2}) \cdot 3 = 0 : u = \frac{- 1 + 73}{12}, u = \frac{- 1 - 73}{12}

u + (- 1 + 2 u^{2}) \cdot 3 = 0

Expand

u + (- 1 + 2 u^{2}) \cdot 3 : u - 3 + 6 u^{2}

u + (- 1 + 2 u^{2}) \cdot 3

= u + 3 (- 1 + 2 u^{2})

Expand

3 (- 1 + 2 u^{2}) : - 3 + 6 u^{2}

3 (- 1 + 2 u^{2})

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = - 1, c = 2 u^{2}

= 3 (- 1) + 3 \cdot 2 u^{2}

Apply minus-plus rules

+ (- a) = - a

= - 3 \cdot 1 + 3 \cdot 2 u^{2}

Simplify

- 3 \cdot 1 + 3 \cdot 2 u^{2} : - 3 + 6 u^{2}

- 3 \cdot 1 + 3 \cdot 2 u^{2}

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 \cdot 2 u^{2}

Multiply the numbers:

3 \cdot 2 = 6

= - 3 + 6 u^{2}

= - 3 + 6 u^{2}

= u - 3 + 6 u^{2}

u - 3 + 6 u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 6, b = 1, c = - 3

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

1^{2} - 4 \cdot 6 (- 3) = 73

1^{2} - 4 \cdot 6 (- 3)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 6 (- 3)

Apply rule

- (- a) = a

= 1 + 4 \cdot 6 \cdot 3

Multiply the numbers:

4 \cdot 6 \cdot 3 = 72

= 1 + 72

Add the numbers:

1 + 72 = 73

= 73

u_{1, 2} = \frac{- 1 \pm 73}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- 1 + 73}{2 \cdot 6}, u_{2} = \frac{- 1 - 73}{2 \cdot 6}

u = \frac{- 1 + 73}{2 \cdot 6} : \frac{- 1 + 73}{12}

\frac{- 1 + 73}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 1 + 73}{12}

u = \frac{- 1 - 73}{2 \cdot 6} : \frac{- 1 - 73}{12}

\frac{- 1 - 73}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 1 - 73}{12}

The solutions to the quadratic equation are:

u = \frac{- 1 + 73}{12}, u = \frac{- 1 - 73}{12}

Substitute back

u = cos (x)

cos (x) = \frac{- 1 + 73}{12}, cos (x) = \frac{- 1 - 73}{12}

cos (x) = \frac{- 1 + 73}{12}, cos (x) = \frac{- 1 - 73}{12}

cos (x) = \frac{- 1 + 73}{12} : x = arccos (\frac{- 1 + 73}{12}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 73}{12}) + 2 πn

cos (x) = \frac{- 1 + 73}{12}

Apply trig inverse properties

cos (x) = \frac{- 1 + 73}{12}

General solutions for

cos (x) = \frac{- 1 + 73}{12}

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

x = arccos (\frac{- 1 + 73}{12}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 73}{12}) + 2 πn

x = arccos (\frac{- 1 + 73}{12}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 73}{12}) + 2 πn

cos (x) = \frac{- 1 - 73}{12} : x = arccos (\frac{- 1 - 73}{12}) + 2 πn, x = - arccos (\frac{- 1 - 73}{12}) + 2 πn

cos (x) = \frac{- 1 - 73}{12}

Apply trig inverse properties

cos (x) = \frac{- 1 - 73}{12}

General solutions for

cos (x) = \frac{- 1 - 73}{12}

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

x = arccos (\frac{- 1 - 73}{12}) + 2 πn, x = - arccos (\frac{- 1 - 73}{12}) + 2 πn

x = arccos (\frac{- 1 - 73}{12}) + 2 πn, x = - arccos (\frac{- 1 - 73}{12}) + 2 πn

Combine all the solutions

x = arccos (\frac{- 1 + 73}{12}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 73}{12}) + 2 πn, x = arccos (\frac{- 1 - 73}{12}) + 2 πn, x = - arccos (\frac{- 1 - 73}{12}) + 2 πn

Show solutions in decimal form

x = 0.89095 \dots + 2 πn, x = 2 π - 0.89095 \dots + 2 πn, x = 2.49035 \dots + 2 πn, x = - 2.49035 \dots + 2 πn

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