What is the general solution for cos(2x)-4cos(x)+3=0 ?

The general solution for cos(2x)-4cos(x)+3=0 is x=2pin

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cos(2x)-4cos(x)+3=0

Solution

cos (2 x) - 4 cos (x) + 3 = 0

Solution

x = 2 πn

Degrees

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

Solution steps

cos (2 x) - 4 cos (x) + 3 = 0

Rewrite using trig identities

3 + cos (2 x) - 4 cos (x)

Use the Double Angle identity:

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

= 3 + 2 cos^{2} (x) - 1 - 4 cos (x)

Simplify

3 + 2 cos^{2} (x) - 1 - 4 cos (x) : 2 cos^{2} (x) - 4 cos (x) + 2

3 + 2 cos^{2} (x) - 1 - 4 cos (x)

Group like terms

= 2 cos^{2} (x) - 4 cos (x) + 3 - 1

Add/Subtract the numbers:

3 - 1 = 2

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

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

2 + 2 cos^{2} (x) - 4 cos (x) = 0

Solve by substitution

2 + 2 cos^{2} (x) - 4 cos (x) = 0

Let:

cos (x) = u

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

2 + 2 u^{2} - 4 u = 0 : u = 1

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 4, c = 2

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

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

(- 4)^{2} - 4 \cdot 2 \cdot 2 = 0

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

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 2}

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

u = 1

The solution to the quadratic equation is:

u = 1

Substitute back

u = cos (x)

cos (x) = 1

cos (x) = 1

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 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}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

Combine all the solutions

x = 2 πn

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2sin(x)+3cot(x)-3csc(x)=0

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cos(2x-pi/6)=-1/2 , pi/2 <= x<pi

cos (2 x - \frac{π}{6}) = - \frac{1}{2}, \frac{π}{2} \leq x < π

Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)-4cos(x)+3=0 ?
The general solution for cos(2x)-4cos(x)+3=0 is x=2pin