What is the general solution for 4cos(2θ)-10cos(θ)+14=7,0<= θ<360 ?

Q: What is the general solution for 4cos(2θ)-10cos(θ)+14=7,0<= θ<360 ?

The general solution for 4cos(2θ)-10cos(θ)+14=7,0<= θ<360 is θ=0.72273…,θ=360-0.72273…,θ=60,θ=300

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4cos(2θ)-10cos(θ)+14=7,0<= θ<360

Solution

4 cos (2 θ) - 10 cos (θ) + 14 = 7, 0 \leq θ < 36 0^{\circ}

Solution

θ = 0.72273 \dots, θ = 36 0^{\circ} - 0.72273 \dots, θ = 6 0^{\circ}, θ = 30 0^{\circ}

Radians

θ = 0.72273 \dots, θ = 2 π - 0.72273 \dots, θ = \frac{π}{3}, θ = \frac{5 π}{3}

Solution steps

4 cos (2 θ) - 10 cos (θ) + 14 = 7, 0 \leq θ < 36 0^{\circ}

Subtract

7

from both sides

4 cos (2 θ) - 10 cos (θ) + 7 = 0

Rewrite using trig identities

7 - 10 cos (θ) + 4 cos (2 θ)

Use the Double Angle identity:

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

= 7 - 10 cos (θ) + 4 (2 cos^{2} (θ) - 1)

Simplify

7 - 10 cos (θ) + 4 (2 cos^{2} (θ) - 1) : 8 cos^{2} (θ) - 10 cos (θ) + 3

7 - 10 cos (θ) + 4 (2 cos^{2} (θ) - 1)

Expand

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

4 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (θ), c = 1

= 4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

Simplify

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

4 \cdot 2 cos^{2} (θ) - 4 \cdot 1

Multiply the numbers:

4 \cdot 2 = 8

= 8 cos^{2} (θ) - 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 8 cos^{2} (θ) - 4

= 8 cos^{2} (θ) - 4

= 7 - 10 cos (θ) + 8 cos^{2} (θ) - 4

Simplify

7 - 10 cos (θ) + 8 cos^{2} (θ) - 4 : 8 cos^{2} (θ) - 10 cos (θ) + 3

7 - 10 cos (θ) + 8 cos^{2} (θ) - 4

Group like terms

= - 10 cos (θ) + 8 cos^{2} (θ) + 7 - 4

Add/Subtract the numbers:

7 - 4 = 3

= 8 cos^{2} (θ) - 10 cos (θ) + 3

= 8 cos^{2} (θ) - 10 cos (θ) + 3

= 8 cos^{2} (θ) - 10 cos (θ) + 3

3 - 10 cos (θ) + 8 cos^{2} (θ) = 0

Solve by substitution

3 - 10 cos (θ) + 8 cos^{2} (θ) = 0

Let:

cos (θ) = u

3 - 10 u + 8 u^{2} = 0

3 - 10 u + 8 u^{2} = 0 : u = \frac{3}{4}, u = \frac{1}{2}

3 - 10 u + 8 u^{2} = 0

Write in the standard form

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

8 u^{2} - 10 u + 3 = 0

Solve with the quadratic formula

8 u^{2} - 10 u + 3 = 0

Quadratic Equation Formula:

For

a = 8, b = - 10, c = 3

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

(- 10)^{2} - 4 \cdot 8 \cdot 3 = 2

(- 10)^{2} - 4 \cdot 8 \cdot 3

Apply exponent rule:

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

n

is even

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 8 \cdot 3

Multiply the numbers:

4 \cdot 8 \cdot 3 = 96

= 1 0^{2} - 96

1 0^{2} = 100

= 100 - 96

Subtract the numbers:

100 - 96 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 10 ) \pm 2}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- ( - 10 ) + 2}{2 \cdot 8}, u_{2} = \frac{- ( - 10 ) - 2}{2 \cdot 8}

u = \frac{- ( - 10 ) + 2}{2 \cdot 8} : \frac{3}{4}

\frac{- ( - 10 ) + 2}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{10 + 2}{2 \cdot 8}

Add the numbers:

10 + 2 = 12

= \frac{12}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{12}{16}

Cancel the common factor:

4

= \frac{3}{4}

u = \frac{- ( - 10 ) - 2}{2 \cdot 8} : \frac{1}{2}

\frac{- ( - 10 ) - 2}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{10 - 2}{2 \cdot 8}

Subtract the numbers:

10 - 2 = 8

= \frac{8}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{8}{16}

Cancel the common factor:

8

= \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{3}{4}, u = \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{3}{4}, cos (θ) = \frac{1}{2}

cos (θ) = \frac{3}{4}, cos (θ) = \frac{1}{2}

cos (θ) = \frac{3}{4}, 0 \leq θ < 36 0^{\circ} : θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4})

cos (θ) = \frac{3}{4}, 0 \leq θ < 36 0^{\circ}

Apply trig inverse properties

cos (θ) = \frac{3}{4}

General solutions for

cos (θ) = \frac{3}{4}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

θ = arccos (\frac{3}{4}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{4}) + 36 0^{\circ} n

θ = arccos (\frac{3}{4}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{4}) + 36 0^{\circ} n

Solutions for the range

0 \leq θ < 36 0^{\circ}

θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4})

cos (θ) = \frac{1}{2}, 0 \leq θ < 36 0^{\circ} : θ = 6 0^{\circ}, θ = 30 0^{\circ}

cos (θ) = \frac{1}{2}, 0 \leq θ < 36 0^{\circ}

General solutions for

cos (θ) = \frac{1}{2}

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq θ < 36 0^{\circ}

θ = 6 0^{\circ}, θ = 30 0^{\circ}

Combine all the solutions

θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4}), θ = 6 0^{\circ}, θ = 30 0^{\circ}

Show solutions in decimal form

θ = 0.72273 \dots, θ = 36 0^{\circ} - 0.72273 \dots, θ = 6 0^{\circ}, θ = 30 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos(2θ)-10cos(θ)+14=7,0<= θ<360 ?
The general solution for 4cos(2θ)-10cos(θ)+14=7,0<= θ<360 is θ=0.72273…,θ=360-0.72273…,θ=60,θ=300