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

The general solution for 4cos^2(x)-4sqrt(2)cos(x)+2=0 is x= pi/4+2pin,x=(7pi)/4+2pin

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4cos^2(x)-4sqrt(2)cos(x)+2=0

Solution

4 cos^{2} (x) - 42 cos (x) + 2 = 0

Solution

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (x) - 42 cos (x) + 2 = 0

Solve by substitution

4 cos^{2} (x) - 42 cos (x) + 2 = 0

Let:

cos (x) = u

4 u^{2} - 42 u + 2 = 0

4 u^{2} - 42 u + 2 = 0 : u = \frac{2}{2}

4 u^{2} - 42 u + 2 = 0

Solve with the quadratic formula

4 u^{2} - 42 u + 2 = 0

Quadratic Equation Formula:

For

a = 4, b = - 42, c = 2

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

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

(- 42)^{2} - 4 \cdot 4 \cdot 2 = 0

(- 42)^{2} - 4 \cdot 4 \cdot 2

(- 42)^{2} = 4^{2} \cdot 2

(- 42)^{2}

Apply exponent rule:

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

n

is even

(- 42)^{2} = (42)^{2}

= (42)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

(2)^{2} : 2

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

2

= 1

= 2

= 4^{2} \cdot 2

4 \cdot 4 \cdot 2 = 32

4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 32

= 4^{2} \cdot 2 - 32

4^{2} \cdot 2 = 32

4^{2} \cdot 2

4^{2} = 16

= 16 \cdot 2

Multiply the numbers:

16 \cdot 2 = 32

= 32

= 32 - 32

Subtract the numbers:

32 - 32 = 0

= 0

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4 2}{8}

Cancel the common factor:

4

= \frac{2}{2}

u = \frac{2}{2}

The solution to the quadratic equation is:

u = \frac{2}{2}

Substitute back

u = cos (x)

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

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

cos (x) = \frac{2}{2} : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

General solutions for

cos (x) = \frac{2}{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}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Popular Examples

sin((2pi)/3-x)+sin((2pi)/3+x)=0

sin (\frac{2 π}{3} - x) + sin (\frac{2 π}{3} + x) = 0

sin(θ)cos(θ)=cot(θ)

sin (θ) cos (θ) = cot (θ)

cot(θ)=-4/3

cot (θ) = - \frac{4}{3}

tan(x)= 6/10

tan (x) = \frac{6}{10}

tan(A)=34

tan (A) = 34

Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-4sqrt(2)cos(x)+2=0 ?
The general solution for 4cos^2(x)-4sqrt(2)cos(x)+2=0 is x= pi/4+2pin,x=(7pi)/4+2pin