What is the general solution for 2cos^2(x)+cos(x)-6=0 ?

The general solution for 2cos^2(x)+cos(x)-6=0 is No Solution for x\in\mathbb{R}

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Popular >

2cos^2(x)+cos(x)-6=0

Solution

2 cos^{2} (x) + cos (x) - 6 = 0

Solution

No Solution for x \in R

Solution steps

2 cos^{2} (x) + cos (x) - 6 = 0

Solve by substitution

2 cos^{2} (x) + cos (x) - 6 = 0

Let:

cos (x) = u

2 u^{2} + u - 6 = 0

2 u^{2} + u - 6 = 0 : u = \frac{3}{2}, u = - 2

2 u^{2} + u - 6 = 0

Solve with the quadratic formula

2 u^{2} + u - 6 = 0

Quadratic Equation Formula:

For

a = 2, b = 1, c = - 6

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 6) = 7

1^{2} - 4 \cdot 2 (- 6)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 6)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 6

Multiply the numbers:

4 \cdot 2 \cdot 6 = 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

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

Separate the solutions

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

u = \frac{- 1 + 7}{2 \cdot 2} : \frac{3}{2}

\frac{- 1 + 7}{2 \cdot 2}

Add/Subtract the numbers:

- 1 + 7 = 6

= \frac{6}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6}{4}

Cancel the common factor:

2

= \frac{3}{2}

u = \frac{- 1 - 7}{2 \cdot 2} : - 2

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

Subtract the numbers:

- 1 - 7 = - 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 8}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

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

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

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cos (a) = \frac{6}{7.5}

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tan (2 x) + sec (2 x) = 4

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)+cos(x)-6=0 ?
The general solution for 2cos^2(x)+cos(x)-6=0 is No Solution for x\in\mathbb{R}