What is the general solution for 4cos^2(x)-4=15cos(x) ?

The general solution for 4cos^2(x)-4=15cos(x) is x=1.82347…+2pin,x=-1.82347…+2pin

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4cos^2(x)-4=15cos(x)

Solution

4 cos^{2} (x) - 4 = 15 cos (x)

Solution

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

Degrees

x = 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = - 104.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (x) - 4 = 15 cos (x)

Solve by substitution

4 cos^{2} (x) - 4 = 15 cos (x)

Let:

cos (x) = u

4 u^{2} - 4 = 15 u

4 u^{2} - 4 = 15 u : u = 4, u = - \frac{1}{4}

4 u^{2} - 4 = 15 u

Move

15 u

to the left side

4 u^{2} - 4 = 15 u

Subtract

15 u

from both sides

4 u^{2} - 4 - 15 u = 15 u - 15 u

Simplify

4 u^{2} - 4 - 15 u = 0

4 u^{2} - 4 - 15 u = 0

Write in the standard form

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

4 u^{2} - 15 u - 4 = 0

Solve with the quadratic formula

4 u^{2} - 15 u - 4 = 0

Quadratic Equation Formula:

For

a = 4, b = - 15, c = - 4

u_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

(- 15)^{2} - 4 \cdot 4 (- 4) = 17

(- 15)^{2} - 4 \cdot 4 (- 4)

Apply rule

- (- a) = a

= (- 15)^{2} + 4 \cdot 4 \cdot 4

Apply exponent rule:

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

n

is even

(- 15)^{2} = 1 5^{2}

= 1 5^{2} + 4 \cdot 4 \cdot 4

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 1 5^{2} + 64

1 5^{2} = 225

= 225 + 64

Add the numbers:

225 + 64 = 289

= 289

Factor the number:

289 = 1 7^{2}

= 1 7^{2}

Apply radical rule:

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- ( - 15 ) \pm 17}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 15 ) + 17}{2 \cdot 4}, u_{2} = \frac{- ( - 15 ) - 17}{2 \cdot 4}

u = \frac{- ( - 15 ) + 17}{2 \cdot 4} : 4

\frac{- ( - 15 ) + 17}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{15 + 17}{2 \cdot 4}

Add the numbers:

15 + 17 = 32

= \frac{32}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{32}{8}

Divide the numbers:

\frac{32}{8} = 4

= 4

u = \frac{- ( - 15 ) - 17}{2 \cdot 4} : - \frac{1}{4}

\frac{- ( - 15 ) - 17}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{15 - 17}{2 \cdot 4}

Subtract the numbers:

15 - 17 = - 2

= \frac{- 2}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 2}{8}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{8}

Cancel the common factor:

2

= - \frac{1}{4}

The solutions to the quadratic equation are:

u = 4, u = - \frac{1}{4}

Substitute back

u = cos (x)

cos (x) = 4, cos (x) = - \frac{1}{4}

cos (x) = 4, cos (x) = - \frac{1}{4}

cos (x) = 4 :

No Solution

cos (x) = 4

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - \frac{1}{4} : x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

cos (x) = - \frac{1}{4}

Apply trig inverse properties

cos (x) = - \frac{1}{4}

General solutions for

cos (x) = - \frac{1}{4}

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

x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

Combine all the solutions

x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

Show solutions in decimal form

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-4=15cos(x) ?
The general solution for 4cos^2(x)-4=15cos(x) is x=1.82347…+2pin,x=-1.82347…+2pin