What is the general solution for 2cos^2(x)-cos(x)+1=0 ?

The general solution for 2cos^2(x)-cos(x)+1=0 is No Solution for x\in\mathbb{R}

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Popular >

2cos^2(x)-cos(x)+1=0

Solution

2 cos^{2} (x) - cos (x) + 1 = 0

Solution

No Solution for x \in R

Solution steps

2 cos^{2} (x) - cos (x) + 1 = 0

Solve by substitution

2 cos^{2} (x) - cos (x) + 1 = 0

Let:

cos (x) = u

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

2 u^{2} - u + 1 = 0 : u = \frac{1}{4} + i \frac{7}{4}, u = \frac{1}{4} - i \frac{7}{4}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

Simplify

(- 1)^{2} - 4 \cdot 2 \cdot 1 : 7 i

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 - 8

Subtract the numbers:

1 - 8 = - 7

= - 7

Apply radical rule:

- a = - 1 a

- 7 = - 1 7

= - 1 7

Apply imaginary number rule:

- 1 = i

= 7 i

u_{1, 2} = \frac{- ( - 1 ) \pm 7 i}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 7 i}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 7 i}{2 \cdot 2}

u = \frac{- ( - 1 ) + 7 i}{2 \cdot 2} : \frac{1}{4} + i \frac{7}{4}

\frac{- ( - 1 ) + 7 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 7 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 + 7 i}{4}

Rewrite

\frac{1 + 7 i}{4}

in standard complex form:

\frac{1}{4} + \frac{7}{4} i

\frac{1 + 7 i}{4}

Apply the fraction rule:

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

\frac{1 + 7 i}{4} = \frac{1}{4} + \frac{7 i}{4}

= \frac{1}{4} + \frac{7 i}{4}

= \frac{1}{4} + \frac{7}{4} i

u = \frac{- ( - 1 ) - 7 i}{2 \cdot 2} : \frac{1}{4} - i \frac{7}{4}

\frac{- ( - 1 ) - 7 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 7 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 - 7 i}{4}

Rewrite

\frac{1 - 7 i}{4}

in standard complex form:

\frac{1}{4} - \frac{7}{4} i

\frac{1 - 7 i}{4}

Apply the fraction rule:

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

\frac{1 - 7 i}{4} = \frac{1}{4} - \frac{7 i}{4}

= \frac{1}{4} - \frac{7 i}{4}

= \frac{1}{4} - \frac{7}{4} i

The solutions to the quadratic equation are:

u = \frac{1}{4} + i \frac{7}{4}, u = \frac{1}{4} - i \frac{7}{4}

Substitute back

u = cos (x)

cos (x) = \frac{1}{4} + i \frac{7}{4}, cos (x) = \frac{1}{4} - i \frac{7}{4}

cos (x) = \frac{1}{4} + i \frac{7}{4}, cos (x) = \frac{1}{4} - i \frac{7}{4}

cos (x) = \frac{1}{4} + i \frac{7}{4} :

No Solution

cos (x) = \frac{1}{4} + i \frac{7}{4}

No Solution

cos (x) = \frac{1}{4} - i \frac{7}{4} :

No Solution

cos (x) = \frac{1}{4} - i \frac{7}{4}

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)-cos(x)+1=0 ?
The general solution for 2cos^2(x)-cos(x)+1=0 is No Solution for x\in\mathbb{R}