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Popular Trigonometry >

4cos(2x)-3cos(x)+1=0

Solution

4 cos (2 x) - 3 cos (x) + 1 = 0

Solution

x = 0.59538 \dots + 2 πn, x = 2 π - 0.59538 \dots + 2 πn, x = 2.04085 \dots + 2 πn, x = - 2.04085 \dots + 2 πn

Degrees

x = 34.11286 \dots^{\circ} + 36 0^{\circ} n, x = 325.88713 \dots^{\circ} + 36 0^{\circ} n, x = 116.93210 \dots^{\circ} + 36 0^{\circ} n, x = - 116.93210 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (2 x) - 3 cos (x) + 1 = 0

Rewrite using trig identities

1 - 3 cos (x) + 4 cos (2 x)

Use the Double Angle identity:

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

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

Simplify

1 - 3 cos (x) + 4 (2 cos^{2} (x) - 1) : 8 cos^{2} (x) - 3 cos (x) - 3

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

Expand

4 (2 cos^{2} (x) - 1) : 8 cos^{2} (x) - 4

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

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (x), c = 1

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

Simplify

4 \cdot 2 cos^{2} (x) - 4 \cdot 1 : 8 cos^{2} (x) - 4

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

Multiply the numbers:

4 \cdot 2 = 8

= 8 cos^{2} (x) - 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 8 cos^{2} (x) - 4

= 8 cos^{2} (x) - 4

= 1 - 3 cos (x) + 8 cos^{2} (x) - 4

Simplify

1 - 3 cos (x) + 8 cos^{2} (x) - 4 : 8 cos^{2} (x) - 3 cos (x) - 3

1 - 3 cos (x) + 8 cos^{2} (x) - 4

Group like terms

= - 3 cos (x) + 8 cos^{2} (x) + 1 - 4

Add/Subtract the numbers:

1 - 4 = - 3

= 8 cos^{2} (x) - 3 cos (x) - 3

= 8 cos^{2} (x) - 3 cos (x) - 3

= 8 cos^{2} (x) - 3 cos (x) - 3

- 3 - 3 cos (x) + 8 cos^{2} (x) = 0

Solve by substitution

- 3 - 3 cos (x) + 8 cos^{2} (x) = 0

Let:

cos (x) = u

- 3 - 3 u + 8 u^{2} = 0

- 3 - 3 u + 8 u^{2} = 0 : u = \frac{3 + 105}{16}, u = \frac{3 - 105}{16}

- 3 - 3 u + 8 u^{2} = 0

Write in the standard form

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

8 u^{2} - 3 u - 3 = 0

Solve with the quadratic formula

8 u^{2} - 3 u - 3 = 0

Quadratic Equation Formula:

For

a = 8, b = - 3, c = - 3

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

(- 3)^{2} - 4 \cdot 8 (- 3) = 105

(- 3)^{2} - 4 \cdot 8 (- 3)

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 8 \cdot 3

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 8 \cdot 3

Multiply the numbers:

4 \cdot 8 \cdot 3 = 96

= 3^{2} + 96

3^{2} = 9

= 9 + 96

Add the numbers:

9 + 96 = 105

= 105

u_{1, 2} = \frac{- ( - 3 ) \pm 105}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 105}{2 \cdot 8}, u_{2} = \frac{- ( - 3 ) - 105}{2 \cdot 8}

u = \frac{- ( - 3 ) + 105}{2 \cdot 8} : \frac{3 + 105}{16}

\frac{- ( - 3 ) + 105}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{3 + 105}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{3 + 105}{16}

u = \frac{- ( - 3 ) - 105}{2 \cdot 8} : \frac{3 - 105}{16}

\frac{- ( - 3 ) - 105}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{3 - 105}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{3 - 105}{16}

The solutions to the quadratic equation are:

u = \frac{3 + 105}{16}, u = \frac{3 - 105}{16}

Substitute back

u = cos (x)

cos (x) = \frac{3 + 105}{16}, cos (x) = \frac{3 - 105}{16}

cos (x) = \frac{3 + 105}{16}, cos (x) = \frac{3 - 105}{16}

cos (x) = \frac{3 + 105}{16} : x = arccos (\frac{3 + 105}{16}) + 2 πn, x = 2 π - arccos (\frac{3 + 105}{16}) + 2 πn

cos (x) = \frac{3 + 105}{16}

Apply trig inverse properties

cos (x) = \frac{3 + 105}{16}

General solutions for

cos (x) = \frac{3 + 105}{16}

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

x = arccos (\frac{3 + 105}{16}) + 2 πn, x = 2 π - arccos (\frac{3 + 105}{16}) + 2 πn

x = arccos (\frac{3 + 105}{16}) + 2 πn, x = 2 π - arccos (\frac{3 + 105}{16}) + 2 πn

cos (x) = \frac{3 - 105}{16} : x = arccos (\frac{3 - 105}{16}) + 2 πn, x = - arccos (\frac{3 - 105}{16}) + 2 πn

cos (x) = \frac{3 - 105}{16}

Apply trig inverse properties

cos (x) = \frac{3 - 105}{16}

General solutions for

cos (x) = \frac{3 - 105}{16}

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

x = arccos (\frac{3 - 105}{16}) + 2 πn, x = - arccos (\frac{3 - 105}{16}) + 2 πn

x = arccos (\frac{3 - 105}{16}) + 2 πn, x = - arccos (\frac{3 - 105}{16}) + 2 πn

Combine all the solutions

x = arccos (\frac{3 + 105}{16}) + 2 πn, x = 2 π - arccos (\frac{3 + 105}{16}) + 2 πn, x = arccos (\frac{3 - 105}{16}) + 2 πn, x = - arccos (\frac{3 - 105}{16}) + 2 πn

Show solutions in decimal form

x = 0.59538 \dots + 2 πn, x = 2 π - 0.59538 \dots + 2 πn, x = 2.04085 \dots + 2 πn, x = - 2.04085 \dots + 2 πn

Graph

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