What is the general solution for cos^2(A)+2=3cos(A) ?

The general solution for cos^2(A)+2=3cos(A) is A=2pin

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Popular >

cos^2(A)+2=3cos(A)

Solution

cos^{2} (A) + 2 = 3 cos (A)

Solution

A = 2 πn

Degrees

A = 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (A) + 2 = 3 cos (A)

Solve by substitution

cos^{2} (A) + 2 = 3 cos (A)

Let:

cos (A) = u

u^{2} + 2 = 3 u

u^{2} + 2 = 3 u : u = 2, u = 1

u^{2} + 2 = 3 u

Move

3 u

to the left side

u^{2} + 2 = 3 u

Subtract

3 u

from both sides

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

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

u = \frac{- ( - 3 ) + 1}{2 \cdot 1} : 2

\frac{- ( - 3 ) + 1}{2 \cdot 1}

Apply rule

- (- a) = a

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

Add the numbers:

3 + 1 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

u = \frac{- ( - 3 ) - 1}{2 \cdot 1} : 1

\frac{- ( - 3 ) - 1}{2 \cdot 1}

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 2, u = 1

Substitute back

u = cos (A)

cos (A) = 2, cos (A) = 1

cos (A) = 2, cos (A) = 1

cos (A) = 2 :

No Solution

cos (A) = 2

- 1 \leq cos (x) \leq 1

No Solution

cos (A) = 1 : A = 2 πn

cos (A) = 1

General solutions for

cos (A) = 1

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}

A = 0 + 2 πn

A = 0 + 2 πn

Solve

A = 0 + 2 πn : A = 2 πn

A = 0 + 2 πn

0 + 2 πn = 2 πn

A = 2 πn

A = 2 πn

Combine all the solutions

A = 2 πn

Popular Examples

sin(θ)=(-5)/6

sin (θ) = \frac{- 5}{6}

(sin(x))/(16)=(sin(31))/(12)

\frac{sin ( x )}{16} = \frac{sin ( 3 1 ^{\circ} )}{12}

sin(A)= 2/3

sin (A) = \frac{2}{3}

3cos(2x+1)=2

3 cos (2 x + 1) = 2

sin(t)= 3/7 ,cos(t)

sin (t) = \frac{3}{7}, cos (t)

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(A)+2=3cos(A) ?
The general solution for cos^2(A)+2=3cos(A) is A=2pin