What is the general solution for 8cos(x)=4cos^2(x)+3 ?

The general solution for 8cos(x)=4cos^2(x)+3 is x= pi/3+2pin,x=(5pi)/3+2pin

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8cos(x)=4cos^2(x)+3

Solution

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

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

8 u = 4 u^{2} + 3

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

8 u = 4 u^{2} + 3

Switch sides

4 u^{2} + 3 = 8 u

Move

8 u

to the left side

4 u^{2} + 3 = 8 u

Subtract

8 u

from both sides

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

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = - 8, c = 3

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

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

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

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

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

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

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 8^{2} - 48

8^{2} = 64

= 64 - 48

Subtract the numbers:

64 - 48 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 4}{2 \cdot 4}, u_{2} = \frac{- ( - 8 ) - 4}{2 \cdot 4}

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

\frac{- ( - 8 ) + 4}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{8 + 4}{2 \cdot 4}

Add the numbers:

8 + 4 = 12

= \frac{12}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{12}{8}

Cancel the common factor:

4

= \frac{3}{2}

u = \frac{- ( - 8 ) - 4}{2 \cdot 4} : \frac{1}{2}

\frac{- ( - 8 ) - 4}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{8 - 4}{2 \cdot 4}

Subtract the numbers:

8 - 4 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = \frac{3}{2} :

No Solution

cos (x) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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3 tan^{2} (3 x) = 1

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Frequently Asked Questions (FAQ)

What is the general solution for 8cos(x)=4cos^2(x)+3 ?
The general solution for 8cos(x)=4cos^2(x)+3 is x= pi/3+2pin,x=(5pi)/3+2pin