What is the general solution for cos(2θ)-sqrt(3)cos(θ)=0,0<= θ<2pi ?

Q: What is the general solution for cos(2θ)-sqrt(3)cos(θ)=0,0<= θ<2pi ?

The general solution for cos(2θ)-sqrt(3)cos(θ)=0,0<= θ<2pi is θ=1.97810…,θ=-1.97810…+2pi

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cos(2θ)-sqrt(3)cos(θ)=0,0<= θ<2pi

Solution

cos (2 θ) - 3 cos (θ) = 0, 0 \leq θ < 2 π

Solution

θ = 1.97810 \dots, θ = - 1.97810 \dots + 2 π

Degrees

θ = 113.33731 \dots^{\circ}, θ = 246.66268 \dots^{\circ}

Solution steps

cos (2 θ) - 3 cos (θ) = 0, 0 \leq θ < 2 π

Rewrite using trig identities

cos (2 θ) - cos (θ) 3

Use the Double Angle identity:

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

= 2 cos^{2} (θ) - 1 - 3 cos (θ)

- 1 + 2 cos^{2} (θ) - cos (θ) 3 = 0

Solve by substitution

- 1 + 2 cos^{2} (θ) - cos (θ) 3 = 0

Let:

cos (θ) = u

- 1 + 2 u^{2} - u 3 = 0

- 1 + 2 u^{2} - u 3 = 0 : u = \frac{3 + 11}{4}, u = \frac{3 - 11}{4}

- 1 + 2 u^{2} - u 3 = 0

Write in the standard form

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

2 u^{2} - 3 u - 1 = 0

Solve with the quadratic formula

2 u^{2} - 3 u - 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 3, c = - 1

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

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

(- 3)^{2} - 4 \cdot 2 (- 1) = 11

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

Apply rule

- (- a) = a

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

(- 3)^{2} = 3

(- 3)^{2}

Apply exponent rule:

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

n

is even

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

= (3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 8

= 3 + 8

Add the numbers:

3 + 8 = 11

= 11

u_{1, 2} = \frac{- ( - 3 ) \pm 11}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 11}{2 \cdot 2}, u_{2} = \frac{- ( - 3 ) - 11}{2 \cdot 2}

u = \frac{- ( - 3 ) + 11}{2 \cdot 2} : \frac{3 + 11}{4}

\frac{- ( - 3 ) + 11}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 + 11}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 + 11}{4}

u = \frac{- ( - 3 ) - 11}{2 \cdot 2} : \frac{3 - 11}{4}

\frac{- ( - 3 ) - 11}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 - 11}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 - 11}{4}

The solutions to the quadratic equation are:

u = \frac{3 + 11}{4}, u = \frac{3 - 11}{4}

Substitute back

u = cos (θ)

cos (θ) = \frac{3 + 11}{4}, cos (θ) = \frac{3 - 11}{4}

cos (θ) = \frac{3 + 11}{4}, cos (θ) = \frac{3 - 11}{4}

cos (θ) = \frac{3 + 11}{4}, 0 \leq θ < 2 π :

No Solution

cos (θ) = \frac{3 + 11}{4}, 0 \leq θ < 2 π

- 1 \leq cos (x) \leq 1

No Solution

cos (θ) = \frac{3 - 11}{4}, 0 \leq θ < 2 π : θ = arccos (\frac{3 - 11}{4}), θ = - arccos (\frac{3 - 11}{4}) + 2 π

cos (θ) = \frac{3 - 11}{4}, 0 \leq θ < 2 π

Apply trig inverse properties

cos (θ) = \frac{3 - 11}{4}

General solutions for

cos (θ) = \frac{3 - 11}{4}

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

θ = arccos (\frac{3 - 11}{4}) + 2 πn, θ = - arccos (\frac{3 - 11}{4}) + 2 πn

θ = arccos (\frac{3 - 11}{4}) + 2 πn, θ = - arccos (\frac{3 - 11}{4}) + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = arccos (\frac{3 - 11}{4}), θ = - arccos (\frac{3 - 11}{4}) + 2 π

Combine all the solutions

θ = arccos (\frac{3 - 11}{4}), θ = - arccos (\frac{3 - 11}{4}) + 2 π

Show solutions in decimal form

θ = 1.97810 \dots, θ = - 1.97810 \dots + 2 π

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2θ)-sqrt(3)cos(θ)=0,0<= θ<2pi ?
The general solution for cos(2θ)-sqrt(3)cos(θ)=0,0<= θ<2pi is θ=1.97810…,θ=-1.97810…+2pi